kadane's algorithm implementation
input = (t=1;n=3;arr={-1,4,5}) gives output 8 but expected output was 9.
#include <iostream>
#include<cmath>
using namespace std;
int Max( int *arr, int n){
int currmax = arr[0];
int globalmax = arr[0];
for(int i = 1; i < n; i++) {
currmax= max(currmax, currmax + arr[i]);
if(currmax > globalmax)
globalmax = currmax;
}
return globalmax;
}
int main() {
int t;
cin >> t;
while(t--){
int n;
cin >> n;
int arr[n];
for(int i = 0; i < n; i++) cin >> arr[i];
cout << Max(arr, n) << " ";
}
}
Change currmax= max(currmax, currmax+arr[i]) to currmax= max(arr[i], currmax+arr[i]) as to compare current value with array value.
In you case -1, 4, 5
Value of currmax will be updated as below:
1. currmax = -1, currmax+arr[1] = 3 (compare btw(4,3))
2. currmax= 4 , currmax+arr[2] = 9 (compare btw(5,9))
3. currmax = 9
Code Below:
#include <iostream>
#include <cmath>
using namespace std;
int Max( int *arr, int n){
int currmax=arr[0];
int globalmax=arr[0];
for(int i=1;i<n;i++) {
currmax= max(arr[i], currmax+arr[i]); // your code line was wrong here check and try
if(currmax>globalmax)
globalmax=currmax;
}
return globalmax;
}
int main() {
int t;
cin>>t;
while(t--){
int n;
cin>>n;
int arr[n];
for(int i=0;i<n;i++) cin>>arr[i];
cout<<Max(arr,n)<<" ";
}
}
Please find the correct Answer for Maximum SubArray (Kadane's Theorem)
C# Solution:
public int MaxSubArray(int[] nums) {
int maxSum = nums[0];
int currSum = 0;
for(int i = 0; i < nums.Length; i++){
currSum = currSum + nums[i];
if(maxSum < currSum){
maxSum = currSum;
}
if(currSum < 0){
currSum = 0;
}
}
return maxSum;
}
enter code here
Related
You have been given an array/list(ARR) of size N. You need to swap every pair of alternate elements in the array/list.
You don't need to print or return anything, just change in the input array itself.
#include <iostream>;
using namespace std;
void printArr(int arr[], int n)
{
for (int i = 0; i < n; i++)
cout << arr[i]<<i;
}
void UpdateArr(int arr[], int n)
{
int i = 0, j = n - 1;
while (i < j) {
int temp = arr[i];
arr[i] = arr[j];
arr[j] = temp;
i += 2;
j -= 2;
}
cout<<' printArr(arr[], n)';
}
int main()
{
int t;
cin>> t;
int n;
cin>> n;
int input[100];
for(int i=0; i<n; i++) {
cin >>input[i];
}
int arr[100] ;
n = sizeof(arr) / sizeof(arr[0]);
UpdateArr(arr, n);
return 0;
}
I'm not sure what are you exactly expecting the output to be (pls edit it and show the expected output) but I think this is what you need to do
#include <iostream>
#include <iomanip>
using namespace std;
void UpdateArray(int Arr[], size_t n) {
for (size_t i = 0; i < n / 2; i++) {
int Holder = Arr[i];
Arr[i] = Arr[~i + n];
Arr[~i + n] = Holder; } }
int main() {
int Arr[7] = { 1,2,3,4,5,6,7 };
UpdateArray(Arr, 7);
for (int i = 0; i < 7; i++) {
std::cout << Arr[i] << "\n"; }
return 0; }
size_t is like an int but it can't go into negative, but it can take bigger positive numbers, you can replace it with int, it shouldn't make a difference.
so we loop through half the array, replacing first items with last, the [~i + n] flips the value to the other side, so like index 4 in a array size of 20 will become 15
I would to solve this problem illustrated in this image using C++
Define f(n) as the sum of the factorials of the digits of n. For example, f(342) = 3! + 4! + 2! = 32.
Define sf(n) as the sum of the digits of f(n). So sf(342) = 3 + 2 = 5.
Define g(i) to be the smallest positive integer n such that sf(n) = i. Though sf(342) is 5, sf(25) is also 5, and it can be verified that g(5) is 25.
Define sg(i) as the sum of the digits of g(i). So sg(5) = 2 + 5 = 7.
Further, it can be verified that g(20) is 267 and ∑ sg(i) for 1 ≤ i ≤ 20 is 156.
What is ∑ sg(i) for 1 ≤ i ≤ 150?
Image:
Here is my approach. My code takes long time running, and it works
#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;
int factorial(int n);
std::vector<int> int_to_vector(int n);
int sum_vector(std::vector<int> v);
int get_smallest_number(std::vector<int> v, int sum, int n);
int sum_sg(int n )
int main() {
/* Enter your code here. Read input from STDIN. Print output to STDOUT */
int q;
int m,n;
int g;
int sum = 0;
std::vector<int> vec;
cin>>q;
if( 1<=q && q<=100000){
std::vector<int> s;
int fact = 0;
int sg = 0;
for ( int i = 0; i < q; i++){
cin>>n>>m;
fact = factorial(n);
s = int_to_vector(fact);
sum = sum_vector(s);
g = get_smallest_number(s, sum, n);
s = int_to_vector(g);
sum = sum_vector(s);
}
}
return 0;
}
int factorial(int n){
if (n==0) return 1;
return factorial(n-1)*n;
}
std::vector<int> int_to_vector(int n){
std::vector<int> numbers;
while(n>0)
{
numbers.push_back(n%10);
n/=10;
}
return numbers;
}
int sum_vector(std::vector<int> v){
int sum=0;
for(int i = 0; i < v.size(); i++){
sum+=v.at(i);
}
return sum;
}
int get_smallest_number(std::vector<int> v, int sum, int n){
int i = 0;
int factoriel = 1;
std::vector<int> vect;
int sum2 = 0;
while( i < n){
factoriel = factorial(i);
vect = int_to_vector(factoriel);
sum2 = sum_vector(vect);
if( sum2 == sum) return i;
i++ ;
}
return n;
}
I think on recursive solutions, but it seems somehow more complex to implement. Are there any solutions using modern C++ and STL?
Here is below a bad solution with less complexity, please any suggestion to have good and fast code rather than this below,
#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;
std::vector<int> factorial {1, 1, 2, 6, 24, 120, 720, 5040, 40320, 362880};
std::vector<int> int_to_vector(int n);
int sum_vector(std::vector<int> v);
int find_g(int i);
int calculat_sum_fact(std::vector<int> v);
int main() {
/* Enter your code here. Read input from STDIN. Print output to STDOUT */
int m,n,q;
std::vector<int> vec;
std::vector<int> s;
std::vector<int> numbers;
cin>>q;
for (int i = 0; i < q; i++){
cin>>n>>m;
numbers.push_back(n);
// cout<< numbers.at(i);
}
for(int k = 0; k < numbers.size(); k++){
int sum = 0;
for (int i = 0; i < numbers.at(k)+1; i++){
s = int_to_vector(find_g(i));
sum += sum_vector(s);
}
cout << sum << "\n";
}
return 0;
}
std::vector<int> int_to_vector(int n){
std::vector<int> numbers;
while(n>0)
{
numbers.push_back(n%10);
n/=10;
}
return numbers;
}
int sum_vector(std::vector<int> v){
int sum=0;
for(int i = 0; i < v.size(); i++){
sum+=v.at(i);
}
return sum;
}
int find_g(int i){
std::vector<int> s ;
int sum = 0;
int j = 0;
//cout<<"sf 1= " << j << endl;
while(true){
s = int_to_vector(j);
sum = calculat_sum_fact(s);
s= int_to_vector(sum);
sum = sum_vector(s);
//cout<<"sf 2= " << j << endl;
if (sum == i) {
//cout<<"sf = " << j << endl;
return j;
//break;
}
//cout<<"sf = " << j << endl;
j++;
}
}
int calculat_sum_fact(std::vector<int> v){
int sum = 0;
for(int i = 0; i < v.size(); i++ ){
sum += factorial.at(v.at(i));
}
return sum;
}
I've been stuck on this problem for a pretty Long while. I always get "Time limit exceeded" when I submit the code.
My solution is to input the items of the array then determine the largest number in the array and diplay it along with the elements following it and so on.
How can I make my algorithm more efficient?
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
int main() {
int T, n;
cin >> T;
while (T--) {
//inputting p[n]
scanf_s("%d", &n);
int* p = new int[n];
for (int i = 0; i < n; i++) {
scanf_s("%d", &p[i]);
}
while (n != 0) {
//*itr = largest element in the array
auto itr = find(p, p + n, *max_element(p, p + n));
int index = distance(p, itr);
for (int i = index; i < n; i++) {
printf("%d\n", p[i]);
}
//deleting element by decreasing n:
n = index;
}
delete[] p;
}
return 0;
}
You solution is O(n^2), too slow.
A O(n) solution is obtained by iteratively calculating the position of the max element until a given index i.
#include <iostream>
#include <vector>
#include <algorithm>
//using namespace std;
int main() {
int T;
std::cin >> T;
while (T--) {
//inputting p[n]
int n;
std::cin >> n;
std::vector<int> p(n);
for (int i = 0; i < n; i++) {
std::cin >> p[i];
}
std::vector<int> max_before(n);
max_before[0] = 0;
for (int i = 1; i < n; ++i) {
if (p[i] > p[max_before[i-1]]) {
max_before[i] = i;
} else {
max_before[i] = max_before[i-1];
}
}
while (n != 0) {
int index = max_before[n-1];
for (int i = index; i < n; i++) {
std::cout << p[i] << " ";
}
//deleting element by decreasing n:
n = index;
}
std::cout << '\n';
}
return 0;
}
Sigabrt runtime error occurs of a fatal error, because of an assert statement not returning true? Or use of excessive memory, I'm not able to figure out what I'm doing wrong here, help me out?
( problem 1343 C on codeforces) link
so here's the code.
#include <iostream>
#include <stdlib.h>
#include<vector>
using namespace std;
int check(int i,vector<int> a) {
if (a[i] > 0) {
return 1;
}
else return 0;
}
int main() {
int t;
cin >> t;
while (t--)
{
long int n;
cin >> n;
vector<int> a(n), b;
for (int i = 0; i < n; i++) {
cin >> a[i];
}
int i = 0;
while (i < n)
{
int max = a[i];
int s = check(i,a);
i++;
while (i<n && check(i,a)== s) {
if (a[i] > max)max = a[i];
i++;
}
b.push_back(max);
}
int s = 0;
for (int k = 0; k< b.size(); k++) {
s += b[i];
}
cout << s << endl;
}
}
I have debugged your code and also the modified code has been accepted for the above question.
Mistakes you made:
1. In the below loop, value at i'th index of vector<int> b is being added to long int s. Instead, b[k] should be added to long int s because the variable being used in the loop is k not i.
for (int k = 0; k< b.size(); k++) {
s += b[i];
}
2. In the question, range of variable n is given as (1 ≤ n ≤ 2.10^5). So, it is safe to use int n instead of long int n. Also, when I submitted my code on codeforces it gave me signed integer overflow error when I used long int n.
3. You need to use long long s instead of long int s because the value of each element of array A lies between (−10^9 ≤ a[i] ≤ 10^9 , ai ≠ 0) and when we add the elements it can easily surpass int and long int ranges.
4. Although, the answer got accepted when I used vector<int> a in the function
int check(int i,vector<int> a) {
if (a[i] > 0) {
return 1;
}
else return 0;
}
But as the user Scheff has said and is correct that it comes with a penalty in space and time, you should use call by reference i.e. vector<int> &a.
Modified Code:
#include <iostream>
#include <stdlib.h>
#include<vector>
using namespace std;
int check(int i, vector<int> &a) {
if (a[i] > 0) {
return 1;
}
else return 0;
}
int main() {
int t;
cin >> t;
while (t--)
{
int n;
cin >> n;
vector<int> a(n), b;
for (int i = 0; i < n; i++) {
cin >> a[i];
}
int i = 0;
while (i < n)
{
int max = a[i];
int s = check(i,a);
i++;
while ((i<n) && (check(i,a)== s)) {
if (a[i] > max)
max = a[i];
i++;
}
b.push_back(max);
}
long long s = 0;
for (int k = 0; k< b.size(); k++) {
s += b[k];
}
cout << s << endl;
}
}
Screenshot of Accepted Answer:
I tried to find the biggest number in every row of a 2D array, and to put it on diametrum. I did that, but I now want to replace that number, not just declare them on diametrum. How do I change the place of numbers for my case? Here's code:
#include<iostream>
using namespace std;
int mac(int b[],int n)
{
int i,max,c=0;
max=b[0];
for(i=0;i<n;i++)
{
if(max<b[i]) {max=b[i];c=i;}
}
return c;
}
int main()
{
int i,j,n;
int c;
int a[10][10],b[10];
cin>>n;
for(i=0;i<n;i++)
{
for(j=0;j<n;j++)
{
cin>>a[i][j];
cin.ignore();
}
}
for(i=0;i<n;i++)
{
for(j=0;j<n;j++)
b[j]=a[i][j];
c=mac(b,n);
/*Place where i should change something*/
a[i][i]=a[i][c];
}
for(i=0;i<n;i++)
{
for(j=0;j<n;j++)
{
cout<<a[i][j];
}
cout<<endl;
}
return 0;
}
You need to do swap. So you have to define one temporary variable to save actual value from [i][i]. Then overwrite position [i][i] with max value and at the end you need place temp value back to position where you found max value.
/*Place where i should change something*/
int temp = a[i][i];
a[i][i] = a[i][c];
a[i][c] = temp;
Assuming that when n = 3, and input is 1 2 3 4 5 6 7 8 9 the output should be 321 465 789
The code should be
#include <iostream>
#include <vector>
using namespace std;
int main() {
vector< vector<int> > a;
size_t n;
cout<<"Enter number of value for N: ";
cin>>n;
a.reserve(n);
for (size_t i = 0; i<n;i++) {
a[i].reserve(n);
for (size_t j =0 ; j<n;j++) {
cin>>a[i][j];
}
}
for (size_t i = 0; i<n;i++) {
size_t index = 0;
int max = INT_MIN;
for (size_t j =0 ; j<n;j++) {
if (a[i][j] >= max) {
max = a[i][j];
index = j;
}
}
a[i][index] = a[i][i];
a[i][i] = max;
}
cout<<"\nOutput\n";
for (size_t i = 0; i<n;i++) {
for (size_t j =0 ; j<n;j++) {
cout<<a[i][j]<<" ";
}
cout<<endl;
}
}