I have an interval (m,n) and there I have to print out all the numbers which have different digits. I wrote this, but it only works for 2 digit numbers. I simply do not know how to make it work for anything but 2 digit numbers. I imagine that, if I added as much for loops as the digits of my number it would work, but the interval(m,n) isn't specified so it has to be something reliable. I've been trying to solve this problem on my own for 6 damn hours and I'm absolutely fed up.
Input 97,113;
Output 97,98,102,103,104,105,106,107,108,109
Numbers 99,100,101,110+ don't get printed, because they have 2 digits that are
the same.
#include<conio.h>
#include<math.h>
#include<stdio.h>
int main()
{
int m,n,test,checker=0;
scanf("%d%d",&m,&n);
if(m>n)
{
int holder=n;
n=m;
m=holder;
}
for(int start=m;start<=n;start++)
{
int itemCount=floor(log10(abs(start)))+1;
int nums[itemCount];
int index=0;
test=start;
do
{
int nextVal = test % 10;
nums[index++]=nextVal;
test = test / 10;
}while(test>0);
for (int i = 0; i < itemCount - 1; i++)
{ // read comment by #nbro
for (int j = i + 1; j < itemCount; j++)
{
if (nums[i] == nums[j])
{
checker++;
}
}
if(checker==0)printf("%d ",start);
}
checker=0;
}
}
Since you tagged this as C++, here is a very simple solution using simple modulus and division in a loop. No conversion to string is done.
#include <iostream>
#include <bitset>
bool is_unique_digits(int num)
{
std::bitset<10> numset = 0;
while (num > 0)
{
// get last digit
int val = num % 10;
// if bit is on, then this digit is unique
if (numset[val])
return false;
// turn bit on and remove last digit from number
numset.set(val);
num /= 10;
}
return true;
}
int main()
{
for (int i = 97; i <= 113; ++i)
{
if (is_unique_digits(i))
std::cout << i << "\n";
}
}
The is_unique_digit function simply takes the number and repeatedly extracts the digits from it by taking the last digit in the number. Then this digit is tested to see if the same digit appears in the bitset. If the number already exists, false is immediately returned.
If the number is not in the bitset, then the bit that corresponds to that digit is turned "on" and the number is divided by 10 (effectively removing the last digit from the number). If the loop completes, then true is returned.
Live Example
As an idea for a design:
print the number to a string, if it isn't a string already;
declare an array of int d[10]; and set it to all zeroes
for each ascii digit c of the string,
if (d[c-'0']==1) return 0; // this digit exists already in the number
else d[c-'0']= 1;
just put if(checker==0)printf("%d ",start); outside of second loop the loop
like this
for (int i = 0; i < itemCount - 1; i++)
{
for (int j = i + 1; j < itemCount; j++)
{
if (nums[i] == nums[j])
{
checker++;
break;
}
}
}
if(checker==0)
printf("%d ",start);
checker=0;
However instead of using two nested for loop you can use count array which is more efficient
to check 1 number, you can do
X=10; //number to analyze
char counts[10]; for int i=0;i<10;i++) counts[i]=0;
char number[10];
sprintf(&number,"%s",X); bool bad=false;
for(int i=0;i<strlen(number);i++)
{
if(++counts[number[i]-'0']>1) {bad=true;break;}
}`
Related
Consistently comparing digits symmetrically to its middle digit. If first number is bigger than the last , first is wining and I have to display it else I display last and that keep until I reach middle digit(this is if I have odd number of digits), if digit don't have anything to be compared with it wins automatically.
For example number is 13257 the answer is 7 5 2.
Another one 583241 the answer is 5 8 3.
For now I am only trying to catch when number of digits is odd. And got stuck.. This is my code. The problem is that this code don't display any numbers, but it compares them in the if statement(I checked while debugging).
#include <iostream>
using namespace std;
int countDigit(int n) {
int count = 0;
while (n != 0) {
count++;
n /= 10;
}
return count;
}
int main() {
int n;
cin >> n;
int middle;
int count = countDigit(n);
if (count % 2 == 0) {
cout<<"No mid digit exsist!!";
}
else {
int lastDigit = n % 10;
middle = (count + 1) / 2;
for (int i = 0; i < middle; i++) {
for (int j = lastDigit; j<middle; j--) {
if (i > j) {
cout << i <<' ';
}
else {
cout << j;
}
}
}
}
return 0;
}
An easier approach towards this, in my opinion, would be using strings. You can check the size of the string. If there are even number of characters, you can just compare the first half characters, with the last half. If there are odd numbers, then do the same just print the middle character.
Here's what I'd do for odd number of digits:
string n;
cin>>n;
int i,j;
for(i=0,j=n.size()-1;i<n.size()/2,j>=(n.size()+1)/2;i++,j--)
{
if(n[i]>n[j]) cout<<n[i]<<" ";
else cout<<n[j]<<" ";
}
cout<<n[n.size()/2]<<endl;
We analyze the requirements and then come up with a design.
If we have a number, consisting of digits, we want to compare "left" values with "right" values. So, start somehow at the left and the right index of digits in a number.
Look at this number: 123456789
Index: 012345678
Length: 9
in C and C++ indices start with 0.
So, what will we do?
Compare index 0 with index 8
Compare index 1 with index 7
Compare index 2 with index 6
Compare index 3 with index 5
Compare index 4 with index 4
So, the index from the left is running up and the index from the right is running down.
We continue as long as the left index is less than or equal the right index. All this can be done in a for or while loop.
It does not matter, wether the number of digits is odd or even.
Of course we also do need functions that return the length of a number and a digit of the number at a given position. But I see that you know already how to write these functions. So, I will not explain it further here.
I show you 3 different examples.
Ultra simple and very verbose. Very inefficient, because we do not have arrays.
Still simple, but more compressed. Very inefficient, because we do not have arrays.
C++ solution, not allowed in your case
Verbose
#include <iostream>
// Get the length of a number
unsigned int length(unsigned long long number) {
unsigned int length = 0;
while (number != 0) {
number /= 10;
++length;
}
return length;
}
// Get a digit at a given index of a number
unsigned int digitAt(unsigned int index, unsigned long long number) {
index = length(number) - index - 1;
unsigned int result = 0;
unsigned int count = 0;
while ((number != 0) && (count <= index)) {
result = number % 10;
number /= 10;
++count;
}
return result;
}
// Test
int main() {
unsigned long long number;
if (std::cin >> number) {
unsigned int indexLeft = 0;
unsigned int indexRight = length(number) - 1;
while (indexLeft <= indexRight) {
if (digitAt(indexLeft, number) > digitAt(indexRight, number)) {
std::cout << digitAt(indexLeft, number);
}
else {
std::cout << digitAt(indexRight, number);
}
++indexLeft;
--indexRight;
}
}
}
Compressed
#include <iostream>
// Get the length of a number
size_t length(unsigned long long number) {
size_t length{};
for (; number; number /= 10) ++length;
return length;
}
// Get a digit at a given index of a number
unsigned int digitAt(size_t index, unsigned long long number) {
index = length(number) - index - 1;
unsigned int result{}, count{};
for (; number and count <= index; ++count, number /= 10)
result = number % 10;
return result;
}
// Test
int main() {
if (unsigned long long number; std::cin >> number) {
// Iterate from left and right at the same time
for (size_t indexLeft{}, indexRight{ length(number) - 1 }; indexLeft <= indexRight; ++indexLeft, --indexRight)
std::cout << ((digitAt(indexLeft,number) > digitAt(indexRight, number)) ? digitAt(indexLeft, number) : digitAt(indexRight, number));
}
}
More modern C++
#include <iostream>
#include <string>
#include <algorithm>
#include <cctype>
int main() {
if (std::string numberAsString{}; std::getline(std::cin, numberAsString) and not numberAsString.empty() and
std::all_of(numberAsString.begin(), numberAsString.end(), std::isdigit)) {
for (size_t indexLeft{}, indexRight{ numberAsString.length() - 1 }; indexLeft <= indexRight; ++indexLeft, --indexRight)
std::cout << ((numberAsString[indexLeft] > numberAsString[indexRight]) ? numberAsString[indexLeft] : numberAsString[indexRight]);
}
}
You are trying to do something confusing with nested for-cycles. This is obviously wrong, because there is nothing “quadratic” (with respect to the number of digits) in the entire task. Also, your code doesn’t seem to contain anything that would determine the highest-order digit.
I would suggest that you start with something very simple: string’ify the number and then iterate over the digits in the string. This is obviously neither elegant nor particularly fast, but it will be a working solution to start with and you can improve it later.
BTW, the sooner you get out of the bad habit of using namespace std; the better. It is an antipattern, please avoid it.
Side note: There is no need to treat odd and even numbers of digits differently. Just let the algorithm compare the middle digit (if it exists) against itself and select it; no big deal. It is a tiny efficiency drawback in exchange for a big code simplicity benefit.
#include <cstdint>
#include <iostream>
#include <string>
using std::size_t;
using std::uint64_t;
uint64_t extract_digits(uint64_t source) {
const std::string digits{std::to_string(source)};
auto i = digits.begin();
auto j = digits.rbegin();
const auto iend = i + (digits.size() + 1) / 2;
uint64_t result{0};
for (; i < iend; ++i, ++j) {
result *= 10;
result += (*i > *j ? *i : *j) - '0';
}
return result;
}
int main() {
uint64_t n;
std::cin >> n;
std::cout << extract_digits(n) << std::endl;
}
If the task disallows the use of strings and arrays, you could try using pure arithmetics by constructing a “digit-inverted” version of the number and then iterating over both numbers using division and modulo. This will (still) have obvious limitations that stem from the data type size, some numbers cannot be inverted properly etc. (Use GNU MP for unlimited integers.)
#include <cstdint>
#include <iostream>
using std::size_t;
using std::uint64_t;
uint64_t extract_digits(uint64_t source) {
uint64_t inverted{0};
size_t count{0};
for (uint64_t div = source; div; div /= 10) {
inverted *= 10;
inverted += div % 10;
++count;
}
count += 1;
count /= 2;
uint64_t result{0};
if (count) for(;;) {
const uint64_t a{source % 10}, b{inverted % 10};
result *= 10;
result += a > b ? a : b;
if (!--count) break;
source /= 10;
inverted /= 10;
}
return result;
}
int main() {
uint64_t n;
std::cin >> n;
std::cout << extract_digits(n) << std::endl;
}
Last but not least, I would strongly suggest that you ask questions after you have something buildable and runnable. Having homework solved by someone else defeats the homework’s purpose.
#include <iostream>
#include<ctime>
#include<cstdlib>
#include<string>
#include<cmath>
using namespace std;
int main()
{
bool cont = false;
string str;
int num, num2;
cin >> str >> num;
int arr[10];
int a = pow(10, num);
int b = pow(10, (num - 1));
srand(static_cast<int>(time(NULL)));
do {
num2 = rand() % (a - b) + b;
int r;
int i = 0;
int cpy = num2;
while (cpy != 0) {
r = cpy % 10;
arr[i] = r;
i++;
cpy = cpy / 10;
}
for (int m = 0; m < num; m++)
{
for (int j = 0; j < m; j++) {
if (m != j) {
if (arr[m] == arr[j]) {
break;
}
else {
cont = true;
}
}
}
}
cout << num2 << endl;
} while (!cont);
return 0;
}
I want to take a number from the user and produce such a random number.
For example, if the user entered 8, an 8-digit random number.This number must be unique, so each number must be different from each other,for example:
user enter 5
random number=11225(invalid so take new number)
random number =12345(valid so output)
To do this, I divided the number into its digits and threw it into the array and checked whether it was unique. The Program takes random numbers from the user and throws them into the array.It's all right until this part.But my function to check if this number is unique using the for loop does not work.
Because you need your digits to be unique, it's easier to guarantee the uniqueness up front and then mix it around. The problem-solving principle at play here is to start where you are the most constrained. For you, it's repeating digits, so we ensure that will never happen. It's a lot easier than verifying if we did or not.
This code example will print the unique number to the screen. If you need to actually store it in an int, then there's extra work to be done.
#include <algorithm>
#include <iostream>
#include <numeric>
#include <random>
#include <vector>
int main() {
std::vector<int> digits(10);
std::iota(digits.begin(), digits.end(), 0);
std::shuffle(digits.begin(), digits.end(), std::mt19937(std::random_device{}()));
int x;
std::cout << "Number: ";
std::cin >> x;
for (auto it = digits.begin(); it != digits.begin() + x; ++it) {
std::cout << *it;
}
std::cout << '\n';
}
A few sample runs:
Number: 7
6253079
Number: 3
893
Number: 6
170352
The vector digits holds the digits 0-9, each only appearing once. I then shuffle them around. And based on the number that's input by the user, I then print the first x single digits.
The one downside to this code is that it's possible for 0 to be the first digit, and that may or may not fit in with your rules. If it doesn't, you'd be restricted to a 9-digit number, and the starting value in std::iota would be 1.
First I'm going to recommend you make better choices in naming your variables. You do this:
bool cont = false;
string str;
int num, num2;
cin >> str >> num;
What are num and num2? Give them better names. Why are you cin >> str? I can't even see how you're using it later. But I presume that num is the number of digits you want.
It's also not at all clear what you're using a and b for. Now, I presume this next bit of code is an attempt to create a number. If you're going to blindly try and then when done, see if it's okay, why are you making this so complicated. Instead of this:
num2 = rand() % (a - b) + b;
int r;
int i = 0;
int cpy = num2;
while (cpy != 0) {
r = cpy % 10;
arr[i] = r;
i++;
cpy = cpy / 10;
}
You can do this:
for(int index = 0; index < numberOfDesiredDigits; ++index) {
arr[index] = rand() % 10;
}
I'm not sure why you went for so much more complicated.
I think this is your code where you validate:
// So you iterate the entire array
for (int m = 0; m < num; m++)
{
// And then you check all the values less than the current spot.
for (int j = 0; j < m; j++) {
// This if not needed as j is always less than m.
if (m != j) {
// This if-else is flawed
if (arr[m] == arr[j]) {
break;
}
else {
cont = true;
}
}
}
}
You're trying to make sure you have no duplicates. You're setting cont == true if the first and second digit are different, and you're breaking as soon as you find a dup. I think you need to rethink that.
bool areAllUnique = true;
for (int m = 1; allAreUnique && m < num; m++) {
for (int j = 0; allAreUnique && j < m; ++j) {
allAreUnique = arr[m] != arr[j];
}
}
As soon as we encounter a duplicate, allAreUnique becomes false and we break out of both for-loops.
Then you can check it.
Note that I also start the first loop at 1 instead of 0. There's no reason to start the outer loop at 0, because then the inner loop becomes a no-op.
A better way is to keep a set of valid digits -- initialized with 1 to 10. Then grab a random number within the size of the set and grabbing the n'th digit from the set and remove it from the set. You'll get a valid result the first time.
I have a program like this: given a sequence of integers, find the biggest prime and its positon.
Example:
input:
9 // how many numbers
19 7 81 33 17 4 19 21 13
output:
19 // the biggest prime
1 7 // and its positon
So first I get the input, store it in an array, make a copy of that array and sort it (because I use a varible to keep track of the higest prime, and insane thing will happen if that was unsorted) work with every number of that array to check if it is prime, loop through it again to have the positon and print the result.
But the time is too slow, can I improve it?
My code:
#include <iostream>
#include <cmath>
#include <algorithm>
using namespace std;
int main()
{
int n;
cin >> n;
int numbersNotSorted[n];
int maxNum{0};
for (int i = 0; i < n; i++)
{
cin >> numbersNotSorted[i];
}
int numbersSorted[n];
for (int i = 0; i < n; i++)
{
numbersSorted[i] = numbersNotSorted[i];
}
sort(numbersSorted, numbersSorted + n);
for (int number = 0; number < n; number++)
{
int countNum{0};
for (int i = 2; i <= sqrt(numbersSorted[number]); i++)
{
if (numbersSorted[number] % i == 0)
countNum++;
}
if (countNum == 0)
{
maxNum = numbersSorted[number];
}
}
cout << maxNum << '\n';
for (int i = 0; i < n; i++)
{
if (numbersNotSorted[i] == maxNum)
cout << i + 1 << ' ';
}
}
If you need the biggest prime, sorting the array brings you no benefit, you'll need to check all the values stored in the array anyway.
Even if you implemented a fast sorting algorithm, the best averages you can hope for are O(N + k), so just sorting the array is actually more costly than looking for the largest prime in an unsorted array.
The process is pretty straight forward, check if the next value is larger than the current largest prime, and if so check if it's also prime, store the positions and/or value if it is, if not, check the next value, repeat until the end of the array.
θ(N) time compexity will be the best optimization possible given the conditions.
Start with a basic "for each number entered" loop:
#include <iostream>
#include <cmath>
#include <algorithm>
using namespace std;
int main() {
int n;
int newNumber;
cin >> n;
for (int i = 0; i < n; i++) {
cin >> newNumber;
}
}
If the new number is smaller than the current largest prime, then it can be ignored.
int main() {
int n;
int newNumber;
int highestPrime;
cin >> n;
for (int i = 0; i < n; i++) {
cin >> newNumber;
if(newNumber >= highestPrime) {
}
}
}
If the new number is equal to the highest prime, then you just need to store its position somewhere. I'm lazy, so:
int main() {
int n;
int newNumber;
int highestPrime;
int maxPositions = 1234;
int positionList[maxPositions];
int nextPosition;
int currentPosition = 0;
cin >> n;
for (int i = 0; i < n; i++) {
cin >> newNumber;
currentPosition++;
if(newNumber >= highestPrime) {
if(newNumber == highestPrime) {
if(nextPosition+1 >= maxPositions) {
// List of positions is too small (should've used malloc/realloc instead of being lazy)!
} else {
positionList[nextPosition++] = currentPosition;
}
}
}
}
}
If the new number is larger than the current largest prime, then you need to figure out if it is a prime number, and if it is you need to reset the list and store its position, etc:
int main() {
int n;
int newNumber;
int highestPrime = 0;
int maxPositions = 1234;
int positionList[maxPositions];
int nextPosition;
int currentPosition = 0;
cin >> n;
for (int i = 0; i < n; i++) {
cin >> newNumber;
currentPosition++;
if(newNumber >= highestPrime) {
if(newNumber == highestPrime) {
if(nextPosition+1 >= maxPositions) {
// List of positions is too small (should've used malloc/realloc instead of being lazy)!
} else {
positionList[nextPosition++] = currentPosition;
}
} else { // newNumber > highestPrime
if(isPrime(newNumber)) {
nextPosition = 0; // Reset the list
highestPrime = newNumber;
positionList[nextPosition++] = currentPosition;
}
}
}
}
}
You'll also want something to display the results:
if(highestPrime > 0) {
for(nextPosition= 0; nextPosition < currentPosition; nextPosition++) {
cout << positionList[nextPosition];
}
}
Now; the only thing you're missing is an isPrime(int n) function. The fastest way to do that is to pre-calculate a "is/isn't prime" bitfield. It might look something like:
bool isPrime(int n) {
if(n & 1 != 0) {
n >>= 1;
if( primeNumberBitfield[n / 32] & (1 << (n % 32)) != 0) {
return true;
}
}
return false;
}
The problem here is that (for positive values in a 32-bit signed integer) you'll need 1 billion bits (or 128 MiB).
To avoid that you can use a much smaller bitfield for numbers up to sqrt(1 << 31) (which is only about 4 KiB); then if the number is too large for the bitfield you can use the bitfield to find prime numbers and check (with modulo) if they divide the original number evenly.
Note that Sieve of Eratosthenes ( https://en.wikipedia.org/wiki/Sieve_of_Eratosthenes ) is an efficient way to generate that smaller bitfield (but is not efficient to use for a sparse population of larger numbers).
If you do it right, you'll probably create the illusion that it's instantaneous because almost all of the work will be done while a human is slowly typing the numbers in (and not left until after all of the numbers have been entered). For a very fast typist you'll have ~2 milliseconds between numbers, and (after the last number is entered) humans can't notice delays smaller than about 10 milliseconds.
But the time is too slow, can I improve it?
Below loop suffers from:
Why check smallest values first? Makes more sense to check largest values first to find the largest prime. Exit the for (... number..) loop early once a prime is found. This takes advantage of the work done by sort().
Once a candidate value is not a prime, quit testing for prime-ness.
.
// (1) Start for other end rather than as below
for (int number = 0; number < n; number++) {
int countNum {0};
for (int i = 2; i <= sqrt(numbersSorted[number]); i++) {
if (numbersSorted[number] % i == 0)
// (2) No point in continuing prime testing, Value is composite.
countNum++;
}
if (countNum == 0) {
maxNum = numbersSorted[number];
}
}
Corrections left for OP to implement.
Advanced: Prime testing is a deep subject and many optimizations (trivial and complex) exist that are better than OP's approach. Yet I suspect the above 2 improvement will suffice for OP.
Brittleness: Code does not well handle the case of no primes in the list or n <= 0.
i <= sqrt(numbersSorted[number]) is prone to FP issues leading to an incorrect results. Recommend i <= numbersSorted[number]/i).
Sorting is O(n * log n). Prime testing, as done here, is O(n * sqrt(n[i])). Sorting does not increase O() of the overall code when the square root of the max value is less than log of n. Sorting is worth doing if the result of the sort is used well.
Code fails if the largest value was 1 as prime test incorrectly identifies 1 as a prime.
Code fails if numbersSorted[number] < 0 due to sqrt().
Simply full-range int prime test:
bool isprime(int num) {
if (num % 2 == 0) return num == 2;
for (int divisor = 3; divisor <= num / divisor; divisor += 2) {
if (num % divisor == 0) return false;
}
return num > 1;
}
If you want to find the prime, don't go for sorting. You'll have to check for all the numbers present in the array then.
You can try this approach to do the same thing, but all within a lesser amount of time:
Step-1: Create a global function for detecting a prime number. Here's how you can approach this-
bool prime(int n)
{
int i, p=1;
for(i=2;i<=sqrt(n);i++) //note that I've iterated till the square root of n, to cut down on the computational time
{
if(n%i==0)
{
p=0;
break;
}
}
if(p==0)
return false;
else
return true;
}
Step-2: Now your main function starts. You take input from the user:
int main()
{
int n, i, MAX;
cout<<"Enter the number of elements: ";
cin>>n;
int arr[n];
cout<<"Enter the array elements: ";
for(i=0;i<n;i++)
cin>>arr[i];
Step-3: Note that I've declared a counter variable MAX. I initialize this variable as the first element of the array: MAX=arr[0];
Step-4: Now the loop for iterating the array. What I did was, I iterated through the array and at each element, I checked if the value is greater than or equal to the previous MAX. This will ensure, that the program does not check the values which are less than MAX, thus eliminating a part of the array and cutting down the time. I then nested another if statement, to check if the value is a prime or not. If both of these are satisfied, I set the value of MAX to the current value of the array:
for(i=0;i<n;i++)
{
if(arr[i]>=MAX) //this will check if the number is greater than the previous MAX number or not
{
if(prime(arr[i])) //if the previous condition satisfies, then only this block of code will run and check if it's a prime or not
MAX=arr[i];
}
}
What happens is this- The value of MAX changes to the max prime number of the array after every single loop.
Step-5: Then, after finally traversing the array, when the program finally comes out of the loop, MAX will have the largest prime number of the array stored in it. Print this value of MAX. Now for getting the positions where MAX happens, just iterate over the whole loop and check for the values that match MAX and print their positions:
for(i=0;i<n;i++)
{
if(arr[i]==MAX)
cout<<i+1<<" ";
}
I ran this code in Dev C++ 5.11 and the compilation time was 0.72s.
I am trying to solve a problem where every letter has a respective number such as a-1,b-2....z-26.
Now given a number, in how many ways can the number be decoded is the question. consider an example where 25114 can be decoded as 'BEAN',‘BEAAD’, ‘YAAD’, ‘YAN’, ‘YKD’ and ‘BEKD’. this could be decoded in 6 ways.
I have written code in c++ but I am getting the wrong answer. Please correct my code.
#include<bits/stdc++.h>
using namespace std;
int total = 0;
int arr[100001];
void func(int start,int end,int factor){
if(start==end)
return;
int j =start;
if(factor==2&&j==end-1)//if j is the last element and factor is 2,accessing j+1 element is illegual
return;
if(factor==2){
if((arr[j]*10+arr[j+1])>26)
return;
else{
total++;
func(start+2,end,1);
func(start+2,end,2);
}
}
else{//factor is 1
total++;
func(start+1,end,1);
func(start+1,end,2);
}
}
int main(){
int n;
cin>>n;
for(int i=0;i<n;i++){
int p;
cin>>p;
arr[i]=p;
}
func(0,n,1);
func(0,n,2);
cout<<total<<endl;
return 0;
}
essentially what my code is doing is that it fixes one number from the given array(using one digit or two digits from the the given array) and recurses until all the combinations are covered. for example considering the above case, I first choose '2' as my first digit and decode it as 'B'(factor = 1) and then choose '25' and decode it as 'E'(factor = 2).
**following are the input and output from the following code
input : 25114
expected output : 6
my output : 15
input : 3333333333(10 digits)
expected output : 1
my output : 10
Based on the original program from the question I suggest to count the encodings when you reach the end only (if(start==end)).
As func will always be called twice with factor=1 and factor=2, I can freely choose either condition for counting.
Here is the modified code:
#include<bits/stdc++.h>
using namespace std;
int total = 0;
int arr[100001];
void func(int start,int end,int factor){
if(start==end) {
if(factor == 1) total++; // count once when reaching the end
return;
}
int j =start;
if((factor==2) && (j==end-1))//if j is the last element and factor is 2,accessing j+1 element is illegal
return;
if(factor==2){
if((arr[j]*10+arr[j+1])>26)
return;
else{
//total++;
func(start+2,end,1);
func(start+2,end,2);
}
}
else{//factor is 1
//total++;
func(start+1,end,1);
func(start+1,end,2);
}
return;
}
int main(){
int n;
cin>>n;
for(int i=0;i<n;i++){
int p;
cin>>p;
arr[i]=p;
}
func(0,n,1);
func(0,n,2);
cout<<total<<endl;
return 0;
}
This calculates the expected results from the example input in the question.
$ echo 5 2 5 1 1 4|./program
6
$ echo 10 3 3 3 3 3 3 3 3 3 3|./program
1
There is room for improvement.
Instead of modifying a global variable I would return the number of combinations from func and add the values in the higher level.
I would also handle the distinction between 2-digit and 1-digit numbers in the called func instead of in the caller.
Something like this pseudo code:
int func(int start, int end)
{
if(remaining length is <2) {
// we reached the end, so this is one combination
return 1;
}
if(two-digit number is >26) {
// only a 1-digit number is possible, count remaining combinations
return func(start+1, end);
}
// both a 1-digit or 2-digit number is possible, add the remaining combinations for both cases
return func(start+1) + func(start+2);
}
Your question is tagged as "dynamic-programming", but it is anything but.
Instead, think about the state space and its boundary conditions:
The empty string has zero encodings;
A single digit has a single encoding;
An n-digit string has as many encodings as an (n-1)-digit substring plus as many encodings as an (n-2)-digit substring if the first two digits are <= 26.
Thus, we can walk the string from back to front and store the intermediate results for reuse:
uint64_t solve(std::vector<int>& digits) {
const int n = digits.size();
std::vector<int> encodings(n+1);
encodings[n] = 1;
for (int i = n-1; i >= 0; i--) {
bool two_digits_fit = (i < n - 1) && (digits[i] * 10 + digits[i+1]) <= 26; // What if digits[i] == 0?
encodings[i] = encodings[i+1] + (two_digits_fit ? encodings[i+2] : 0);
}
return encodings[0];
}
I was solving a question, with which I am having some problems:
Complete the function getEqualSumSubstring, which takes a single argument. The single argument is a string s, which contains only non-zero digits.
This function should print the length of longest contiguous substring of s, such that the length of the substring is 2*N digits and the sum of the leftmost N digits is equal to the sum of the rightmost N digits. If there is no such string, your function should print 0.
int getEqualSumSubstring(string s) {
int i=0,j=i,foundLength=0;
for(i=0;i<s.length();i++)
{
for(j=i;j<s.length();j++)
{
int temp = j-i;
if(temp%2==0)
{
int leftSum=0,rightSum=0;
string tempString=s.substr(i,temp);
for(int k=0;k<temp/2;k++)
{
leftSum=leftSum+tempString[k]-'0';
rightSum=rightSum+tempString[k+(temp/2)]-'0';
}
if((leftSum==rightSum)&&(leftSum!=0))
if(s.length()>foundLength)
foundLength=s.length();
}
}
}
return(foundLength);
}
The problem is that this code is working for some samples and not for the others. Since this is an exam type question I don't have the test cases either.
This code works
int getEqualSumSubstring(string s) {
int i=0,j=i,foundLength=0;
for(i=0;i<s.length();i++)
{
for(j=i;j<s.length();j++)
{
int temp = j-i+1;
if(temp%2==0)
{
int leftSum=0,rightSum=0;
string tempString=s.substr(i,temp);
// printf("%d ",tempString.length());
for(int k=0;k<temp/2;k++)
{
leftSum=leftSum+tempString[k]-48;
rightSum=rightSum+tempString[k+(temp/2)]-48;
}
if((leftSum==rightSum)&&(leftSum!=0))
if(tempString.length()>foundLength)
foundLength=tempString.length();
}
}
}
return(foundLength);
}
The temp variable must be j-i+1. Otherwise the case where the whole string is the answer will not be covered. Also, we need to make the change suggested by Scott.
Here's my solution that I can confirm works. The ones above didn't really work for me - they gave me compile errors somehow. I got the same question on InterviewStreet, came up with a bad, incomplete solution that worked for 9/15 of the test cases, so I had to spend some more time coding afterwards.
The idea is that instead of caring about getting the left and right sums (which is what I initially did as well), I will get all the possible substrings out of each half (left and right half) of the given input, sort and append them to two separate lists, and then see if there are any matches.
Why?
Say the strings "423" and "234" have the same sum; if I sorted them, they would both be "234" and thus match. Since these numbers have to be consecutive and equal length, I no longer need to worry about having to add them up as numbers and check.
So, for example, if I'm given 12345678, then on the left side, the for-loop will give me:
[1,12,123,1234,2,23,234,3,34]
And on the right:
[5,56,567,5678,...]
And so forth.
However, I'm only taking substrings of a length of at least 2 into account.
I append each of these substrings, sorted by converting into a character array then converting back into a string, into ArrayLists.
So now that all this is done, the next step is to see if there are identical strings of the same numbers in these two ArrayLists. I simply check each of temp_b's strings against temp_a's first string, then against temp_a's second string, and so forth.
If I get a match (say, "234" and "234"), I'll set the length of those matching substrings as my tempCount (tempCount = 3). I also have another variable called 'count' to keep track of the greatest length of these matching substrings (if this was the first occurrence of a match, then count = 0 is overwritten by tempCount = 3, so count = 3).
As for the odd/even string length with the variable int end, the reason for this is because in the line of code s.length()/2+j, is the length of the input happened to be 11, then:
s.length() = 11
s.length()/2 = 11/5 = 5.5 = 5
So in the for-loop, s.length()/2 + j, where j maxes out at s.length()/2, would become:
5 + 5 = 10
Which falls short of the s.length() that I need to reach for to get the string's last index.
This is because the substring function requires an end index of one greater than what you'd put for something like charAt(i).
Just to demonstrate, an input of "47582139875" will generate the following output:
[47, 457, 4578, 24578, 57, 578, 2578, 58, 258, 28] <-- substrings from left half
[139, 1389, 13789, 135789, 389, 3789, 35789, 789, 5789, 578] <-- substrings from right half
578 <-- the longest one that matched
6 <-- the length of '578' x 2
public static int getEqualSumSubtring(String s){
// run through all possible length combinations of the number string on left and right half
// append sorted versions of these into new ArrayList
ArrayList<String> temp_a = new ArrayList<String>();
ArrayList<String> temp_b = new ArrayList<String>();
int end; // s.length()/2 is an integer that rounds down if length is odd, account for this later
for( int i=0; i<=s.length()/2; i++ ){
for( int j=i; j<=s.length()/2; j++ ){
// only account for substrings with a length of 2 or greater
if( j-i > 1 ){
char[] tempArr1 = s.substring(i,j).toCharArray();
Arrays.sort(tempArr1);
String sorted1 = new String(tempArr1);
temp_a.add(sorted1);
//System.out.println(sorted1);
if( s.length() % 2 == 0 )
end = s.length()/2+j;
else // odd length so we need the extra +1 at the end
end = s.length()/2+j+1;
char[] tempArr2 = s.substring(i+s.length()/2, end).toCharArray();
Arrays.sort(tempArr2);
String sorted2 = new String(tempArr2);
temp_b.add(sorted2);
//System.out.println(sorted2);
}
}
}
// For reference
System.out.println(temp_a);
System.out.println(temp_b);
// If the substrings match, it means they have the same sum
// Keep track of longest substring
int tempCount = 0 ;
int count = 0;
String longestSubstring = "";
for( int i=0; i<temp_a.size(); i++){
for( int j=0; j<temp_b.size(); j++ ){
if( temp_a.get(i).equals(temp_b.get(j)) ){
tempCount = temp_a.get(i).length();
if( tempCount > count ){
count = tempCount;
longestSubstring = temp_a.get(i);
}
}
}
}
System.out.println(longestSubstring);
return count*2;
}
Heres my solution to this question including tests. I've added an extra function just because I feel it makes the solution way easier to read than the solutions above.
#include <string>
#include <iostream>
using namespace std;
int getMaxLenSumSubstring( string s )
{
int N = 0; // The optimal so far...
int leftSum = 0, rightSum=0, strLen=s.size();
int left, right;
for(int i=0;i<strLen/2+1;i++) {
left=(s[i]-int('0')); right=(s[strLen-i-1]-int('0'));
leftSum+=left; rightSum+=right;
if(leftSum==rightSum) N=i+1;
}
return N*2;
}
int getEqualSumSubstring( string s ) {
int maxLen = 0, substrLen, j=1;
for( int i=0;i<s.length();i++ ) {
for( int j=1; j<s.length()-i; j++ ) {
//cout<<"Substring = "<<s.substr(i,j);
substrLen = getMaxLenSumSubstring(s.substr(i,j));
//cout<<", Len ="<<substrLen;
if(substrLen>maxLen) maxLen=substrLen;
}
}
return maxLen;
}
Here are a few tests I ran. Based upon the examples above they seem right.
int main() {
cout<<endl<<"Test 1 :"<<getEqualSumSubstring(string("123231"))<<endl;
cout<<endl<<"Test 2 :"<<getEqualSumSubstring(string("986561517416921217551395112859219257312"))<<endl;
cout<<endl<<"Test 3:"<<getEqualSumSubstring(string("47582139875"))<<endl;
}
Shouldn't the following code use tempString.length() instead of s.length()
if((leftSum==rightSum)&&(leftSum!=0))
if(s.length()>foundLength)
foundLength=s.length();
Below is my code for the question... Thanks !!
public class IntCompl {
public String getEqualSumSubstring_com(String s)
{
int j;
int num=0;
int sum = 0;
int m=s.length();
//calculate String array Length
for (int i=m;i>1;i--)
{
sum = sum + m;
m=m-1;
}
String [] d = new String[sum];
int k=0;
String ans = "NULL";
//Extract strings
for (int i=0;i<s.length()-1;i++)
{
for (j=s.length();j>=i+1;k++,j--)
{
num = k;
d[k] = s.substring(i,j);
}
k=num+1;
}
//Sort strings in such a way that the longest strings precede...
for (int i=0; i<d.length-1; i++)
{
for (int h=1;h<d.length;h++)
{
if (d[i].length() > d[h].length())
{
String temp;
temp=d[i];
d[i]=d[h];
d[h]=temp;
}
}
}
// Look for the Strings with array size 2*N (length in even number) and such that the
//the sum of left N numbers is = to the sum of right N numbers.
//As the strings are already in decending order, longest string is searched first and break the for loop once the string is found.
for (int x=0;x<d.length;x++)
{
int sum1=0,sum2=0;
if (d[x].length()%2==0 && d[x].length()<49)
{
int n;
n = d[x].length()/2;
for (int y=0;y<n;y++)
{
sum1 = sum1 + d[x].charAt(y)-'0';
}
for (int y=n;y<d[x].length();y++)
{
sum2 = sum2 + d[x].charAt(y)-'0';
}
if (sum1==sum2)
{
ans = d[x];
break;
}
}
}
return ans;
}
}
Here is the complete Java Program for this question.
Complexity is O(n^3)
This can however be solved in O(n^2).For O(n^2) complexity solution refer to this link
import java.util.Scanner;
import static java.lang.System.out;
public class SubStringProblem{
public static void main(String args[]){
Scanner sc = new Scanner(System.in);
out.println("Enter the Digit String:");
String s = sc.nextLine();
int n = (new SubStringProblem()).getEqualSumSubString(s);
out.println("The longest Sum SubString is "+n);
}
public int getEqualSumSubString(String s){
int N;
if(s.length()%2==0)
{
//String is even
N = s.length();
}
else{
//String is odd
N=s.length()-1;
}
boolean flag =false;
int sum1,sum2;
do{
for(int k=0;k<=s.length()-N;k++){
sum1=0;
sum2=0;
for(int i =k,j=k+N-1;i<j;i++,j--)
{
sum1=sum1 + Integer.parseInt(s.substring(i,i+1));
sum2+=Integer.parseInt(s.substring(j,j+1));
}
if(sum1==sum2){
return N;
}
}
N-=2;
flag =true;
}while(N>1);
return -1;
}
}
What is your rationale for the number 48 on these two lines?
for(int k=0;k<temp/2;k++)
{
leftSum=leftSum+tempString[k]-48;
rightSum=rightSum+tempString[k+(temp/2)]-48;
}
I am just overly curious and would like to hear the reasoning behind it, because I have a similar solution, but without the 48 and it still works. However, I added the 48 an still got the correct answer.
Simple solution. O(n*n). s - input string.
var longest = 0;
for (var i = 0; i < s.length-1; i++) {
var leftSum = rightSum = 0;
for (var j = i, k = i+1, l = 2; j >=0 && k < s.length; j--, k++, l+=2) {
leftSum += parseInt(s[j]);
rightSum += parseInt(s[k]);
if (leftSum == rightSum && l > longest) {
longest = l;
}
}
}
console.log(longest);