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I have a problem and I have no idea how to start, here it is:
You are given a permutation of 1,2,...,n.
You will be given q queries, each query being one of 2 types:
Type 1: swap elements at positions i and j
Type 2: given a position i, print the length of the longest subarray containing the ith element such that the sum of its elements doesn't exceed n
Input
The first line contains 2 integers n and q (1≤n,q≤105).
The next line contains the n integers of the permutation p1,…,pn (1≤pi≤n).
The next q lines contain queries in the format:
Type 1: 1 i j (1≤i,j≤n)
Type 2: 2 i (1≤i≤n)
Output
For each query of type 2, print the required answer.
Here is my code so far, but it gives me the wrong answer on the last few cases:
#include <iostream>
#include <algorithm>
long long atMostSum(int arr[], int n, int k, int a)
{
int sum = arr[a];
int cnt = 1, maxcnt = 1;
for (int i = 0; i < n; i++) {
if ((sum + arr[i]) <= k) {
sum += arr[i];
cnt++;
}
else if(sum!=0)
{
sum = sum - arr[i - cnt] + arr[i];
}
maxcnt = std::max(cnt, maxcnt);
}
return maxcnt - 1;
}
int main(void) {
int n, q;
std::cin >> n >> q;
int p[n];
for(int i = 0; i < n ; i++) {
std::cin >> p[i];
}
for (int i = 0; i < n; i ++) {
int a;
std::cin >> a;
if (a == 2) {
int m;
std::cin >> m;
std::cout << atMostSum(p, n, n, m) << "\n";
} else {
int l, f;
std::cin >> l >> f;
int temp;
temp = p[l];
p[l] = p[f];
p[f] = temp;
}
}
}
Sample Input
3 3
1 2 3
2 1
2 2
2 3
Sample Output
2
2
1
long long atMostSum(int arr[], int n, int k, int a){
long long ans = arr[a];
for (int i = 0; i <= a; i++) {
long long sum = 0;
for (int j = i; j < n; j++) {
sum += arr[j];
if(sum > n) break;
if(i <= a && a <= j) {
ans = max(ans, sum);
}
}
}
return ans;
}
this isn't best for complexity but should be right...
Cannot provide you with the complete solution, but something to get you started with. You are basically looking for a subset in arr[1...n] which consists of element arr[i] and a sum not exceeding n. So simply look for a sub-array with sum (n - arr[i]) and then it becomes a classical problem of finding a subarray with given sum
You can refer the later part of solution here - https://www.geeksforgeeks.org/find-subarray-with-given-sum/
Related
I'm very new to C++ or even coding. I was trying to make a simple array sorter, where the I first input the number of elements that will be in the array and then input the elements. My outcome should be the array sorted in ascending order. I have not thought about the case if elements inserted are same. So I would love to get some help from you folks.
The main error that I'm facing is that only the first unsorted element is sorted while the rest are either interchanged or left the same.
int main(){
int x;
cout<<"Enter no. of elements"<<endl;
cin>>x;
int A[x];
for (int i = 0;i<x;i++){
cin>>A[i];
}
for(int i=0;i<x;i++)
cout<<A[i]<<",";
int count=0;
if(count <= (x-1)){
for (int i=0;i<(x-1);i++){
if(A[i]>A[i+1]){
int a;
a = A[i];
A[i] = A[(i+1)];
A[i+1] = a;
}
else if(A[i]<A[i+1])
count++;
}
}
cout<<"Sorted array:";
for(int i=0;i<x;i++)
cout<<A[i]<<",";
return 0;
}
You declared a variable length array
int x;
cout<<"Enter no. of elements"<<endl;
cin>>x;
int A[x];
because its size is not a compile-time constant.
However variable length arrays are not a standard C++ feature though some compilers have their own language extensions that support variable length arrays,
It is better to use the class template std::vector.
Another problem is that it seems you are trying to use the bubble sort method to sort the array. But this method requires two loops.
Here is a demonstration program that shows how the bubble sort algorithm can be implemented.
#include <iostream>
int main()
{
int a[] = { 9, 8, 7, 6, 5, 4, 3, 2, 1, 0 };
const size_t N = sizeof( a ) / sizeof( *a );
for (const auto &item : a)
{
std::cout << item << ' ';
}
std::cout << '\n';
for (size_t last = N, sorted = N; not ( last < 2 ); last = sorted)
{
for (size_t i = sorted = 1; i < last; i++)
{
if (a[i] < a[i - 1])
{
// std::swap( a[i-1], a[i] );
int tmp = a[i - 1];
a[i - 1] = a[i];
a[i] = tmp;
sorted = i;
}
}
}
for (const auto &item : a)
{
std::cout << item << ' ';
}
std::cout << '\n';
}
The program output is
9 8 7 6 5 4 3 2 1 0
0 1 2 3 4 5 6 7 8 9
Let us try the following method:
find the largest element in the array and move it to the end, by swapping with the last element;
repeat with the array but the last element, and so on.
To find the largest element in A[0..m-1], scan the array and keep an index to the largest so far, let l. This index can be initialized to 0.
// Move the largest to the end
int l= 0;
for (int i= 1; i < m; i++)
{
if (A[i] > A[l]) l= i;
}
// A[l] is the largest in A[0..m-1]
Swap(A[l], A[m-1]);
// A[m-1] is the largest in A[0..m-1]
To sort, repeat with decreasing m. You can stop when the subarray just holds one element:
// Sort
for (int m= n-1; m > 1; m--)
{
// Move the largest to the end
....
}
Writing the Swap operation and assembling the whole code is your task. Also check
correctness of the Move for the limit cases m= 0, 1, 2.
correctness of the Sort for the limit cases n= 1, 2, 3.
how you could instrument the code to verify that the Move does its job.
how you could instrument the code to verify that the Sort does its job.
what happens in case of equal keys.
Your code can be fixed a bit to make it working.
Just replace if (count <= (x - 1)) with while (count < (x - 1)) and also set count = 0; at start of loop, plus replace else if (A[i] < A[i + 1]) with just else. And your code becomes working!
Necessary fixes I did in code below. Also I did formatting (indents and spaces) to make code looks nicer. Rest remains same.
As I see you have a kind of Bubble Sort.
Try it online!
#include <iostream>
using namespace std;
int main() {
int x;
cout << "Enter no. of elements" << endl;
cin >> x;
int A[x];
for (int i = 0; i < x; i++) {
cin >> A[i];
}
for (int i = 0; i < x; i++)
cout << A[i] << ",";
int count = 0;
while (count < (x - 1)) {
count = 0;
for (int i = 0; i < (x - 1); i++) {
if (A[i] > A[i + 1]) {
int a;
a = A[i];
A[i] = A[(i + 1)];
A[i + 1] = a;
} else
count++;
}
}
cout << "Sorted array:";
for (int i = 0; i < x; i++)
cout << A[i] << ",";
return 0;
}
Input:
10
7 3 5 9 1 8 6 0 2 4
Output:
7,3,5,9,1,8,6,0,2,4,Sorted array:0,1,2,3,4,5,6,7,8,9,
If you are taking the size of array as input from user you have to create your array dynamically in c++ like
int *array=new int(x)
and after taking the inputs of the elements just run a nested loop from 0 to size and
the inner loop from 0 to size-1 and check if(A[i]>A[i+1]) if true then swap the values else continue
hey guys I am trying to solve a problem that requires making a program that searches for number k in n strings in an array and all its previous numbers including zero and finally calculates how many strings in array have these numbers . for example if the input is 2 strings ("0123","012") and search for number 1 the output should be 2 in this case .
so I made an array of strings and 2 loops to search in every char in every string (every element in array) but my program gives me wrong answer I don't know why , am I using a wrong function to search (find function) or what ?
#include <bits/stdc++.h>
using namespace std;
main() {
int n, k, sum = 0, good = 0;
cin >> n >> k;
string x[n];
for (int i = 0; i < n; i++) cin >> x[i];
for (int i = 0; i < n; i++) // string loop
{
for (int m = 0; m <= k; m++) // char loop
{
char c = '0' + m;
size_t search = x[i].find(c);
if (search != string::npos) {
sum++;
}
}
if (sum == (k + 1)) good++;
}
cout << good;
}
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Edit the question to include desired behavior, a specific problem or error, and the shortest code necessary to reproduce the problem. This will help others answer the question.
Closed 3 years ago.
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Given N three-digit numbers, your task is to find bit score of all N numbers and then print the number of pairs possible based on these calculated bit score.
Rule for calculating bit score from three digit number:
From the 3-digit number,
· extract largest digit and multiply by 11 then
· extract smallest digit multiply by 7 then
· add both the result for getting bit pairs.
Note: - Bit score should be of 2-digits, if above results in a 3-digit bit score, simply ignore most significant digit.
Consider following examples:
Say, number is 286
Largest digit is 8 and smallest digit is 2
So, 8*11+2*7 =102 so ignore most significant bit , So bit score = 02.
Say, Number is 123
Largest digit is 3 and smallest digit is 1
So, 3*11+7*1=40, so bit score is 40.
Rules for making pairs from above calculated bit scores
Condition for making pairs are
· Both bit scores should be in either odd position or even position to be eligible to form a pair.
· Pairs can be only made if most significant digit are same and at most two pair can be made for a given significant digit.
Constraints
N<=500
Input Format
First line contains an integer N, denoting the count of numbers.
Second line contains N 3-digit integers delimited by space
Output
One integer value denoting the number of bit pairs.
Test Case
Explanation
Example 1
Input
8 234 567 321 345 123 110 767 111
Output
3
Explanation
After getting the most and least significant digits of the numbers and applying the formula given in Rule 1 we get the bit scores of the numbers as:
58 12 40 76 40 11 19 18
No. of pair possible are 3:
40 appears twice at odd-indices 3 and 5 respectively. Hence, this is one pair.
12, 11, 18 are at even-indices. Hence, two pairs are possible from these three-bit scores.
Hence total pairs possible is 3
#include <iostream>
#include <vector>
using namespace std;
vector<int> correctBitScores(vector<int>);
vector<int> bitScore(vector<int>);
int findPairs(vector<int>);
int main() {
int a, b;
int pairs = 0;
vector<int> vec;
vector<int> bitscore;
cout << "\nEnter count of nos: ";
cin >> a;
for (int i = 0; i < a; i++) {
cin >> b;
vec.push_back(b);
}
bitscore = bitScore(vec);
pairs = findPairs(bitscore);
cout << "Max pairs = " << pairs;
return 0;
}
vector<int> correctBitScores(vector<int> bis) {
int temp = 0;
for (size_t i = 0; i < bis.size(); i++) {
temp = bis[i];
int count = 0;
while (temp > 0) {
temp = temp / 10;
count++;
}
if (count > 2)
bis[i] = abs(100 - bis[i]);
}
/*cout << "\nCorrected" << endl;
for (int i = 0; i < size(bis); i++) {
cout << bis[i] << endl;
}*/
return bis;
}
int findPairs(vector<int> vec) {
int count = 0;
vector<int> odd;
vector<int> even;
for (size_t i = 0; i < vec.size(); i++)
(i % 2 == 0 ? even.push_back(vec[i]) : odd.push_back(vec[i]));
for (size_t j = 0; j < odd.size(); j++)
for (size_t k = j + 1; k < odd.size(); k++) {
if (odd[j] / 10 == odd[k] / 10) {
count++;
odd.erase(odd.begin()+j);
}
}
for (size_t j = 0; j < even.size(); j++)
for (size_t k = j + 1; k < even.size(); k++) {
if (even[j] / 10 == even[k] / 10) {
count++;
even.erase(even.begin() + j);
}
}
return count;
}
vector<int> bitScore(vector<int> v) {
int temp = 0, rem = 0;
vector<int> bs;
for (size_t i = 0; i < v.size(); i++) {
int max = 0, min = 9;
temp = v[i];
while (temp > 0) {
rem = temp % 10;
if (min > rem)
min = rem;
if (max < rem)
max = rem;
temp = temp / 10;
}
int bscore = (max * 11) + (min * 7);
bs.push_back(bscore);
}
/*cout << "\nBit Scores = " << endl;
for (int i = 0; i < size(bs); i++) {
cout << bs[i] << endl;
}*/
bs = correctBitScores(bs);
return bs;
}
I tried doing it very simple c++ code as per my understanding of Que,can you just verify it more test cases.
#include <bits/stdc++.h>
using namespace std;
int main() {
int n,count=0;
cin>>n;
vector<int>v(n);
for(int i=0;i<n;i++){
cin>>v[i];
string s = to_string(v[i]);
sort(s.begin(),s.end());
int temp = (s[s.length()-1]-'0')*11 + (s[0] - '0')*7;
v[i] = temp%100;
}
unordered_map<int ,vector<int>>o,e;
for(int i=0;i<n;i=i+2){
o[v[i]/10].push_back(i+1);
}
for(int i=1;i<n;i=i+2){
e[v[i]/10].push_back(i+1);
}
count=0;
for(int i=0;i<10;i++){
int os=o[i].size(),es=e[i].size();
if(os==2)
count++;
if(es == 2)
count++;
if(os>2 || es>2)
count += 2;
}
cout<<count;
}
I have the folowing problem:
Farmer John has built a new long barn, with N (2 <= N <= 100,000)
stalls. The stalls are located along a straight line at positions
x1,...,xN (0 <= xi <= 1,000,000,000).
His C (2 <= C <= N) cows don't like this barn layout and become
aggressive towards each other once put into a stall. To prevent the
cows from hurting each other, FJ wants to assign the cows to the
stalls, such that the minimum distance between any two of them is as
large as possible. What is the largest minimum distance?
Input
t – the number of test cases, then t test cases follows.
* Line 1: Two space-separated integers: N and C
* Lines 2..N+1: Line i+1 contains an integer stall location, xi
Output
For each test case output one integer: the largest minimum distance.
Example
Input:
1
5 3
1
2
8
4
9
Output:
3 Output details:
FJ can put his 3 cows in the stalls at positions 1, 4 and 8,
resulting in a minimum distance of 3. Submit solution!
My approach was to pick some pair which has a certain gap and check if there are enough elements in the array to satisfy the need of all the cows.
To find these elements,I used binary search.
When I find an element , I reset my left to mid so I can continue based on the number of cows left.
My code:
#include <iostream>
int bsearch(int arr[],int l,int r,int gap , int n,int c){
int stat = 0;
for (int i = 1;i <= c; ++i) {
while(l <= r) {
int mid = (l+r)/2;
int x = n+(i*gap);
if (arr[mid] > x && arr[mid-1] < x) {
l = mid;
++stat;
break;
}
if(arr[mid] < x) {
l = mid + 1;
}
if (arr[mid] > x) {
r = mid - 1;
}
}
}
if (stat == c) {
return 0;
}
else {
return -1;
}
}
int calc(int arr[],int n , int c) {
int max = 0;
for (int i = 0; i < n; ++i) {
for (int j = i+1;j < n; ++j) {
int gap = arr[j] - arr[i];
if (gap > max) {
if (bsearch(arr,j,n-1,gap,arr[j],c) == 0) {
max = gap;
}
}
}
}
return max;
}
using namespace std;
int main() {
ios::sync_with_stdio(0);
cin.tie(0);
int t , n ,c;
cin >> t;
for(int i = 0 ; i < t; ++i) {
cin >> n >> c;
int arr[n];
for (int z = 0 ; z < n; ++z) {
cin >> arr[z];
}
sort(arr,arr+n);
//Output;
int ans = calc(arr,n,c);
cout << ans;
}
return 0;
}
Problem Page:
https://www.spoj.com/problems/AGGRCOW/
Given an array "s" of "n" items, you have for each item an left value "L[i]" and right value "R[i]" and its strength "S[i]",if you pick an element you can not pick L[i] elements on immediate left of it and R[i] on immediate right of it, find the maximum strength possible.
Example input:
5 //n
1 3 7 3 7 //strength
0 0 2 2 2 //Left Value
3 0 1 0 0 //Right Value
Output:
10
Code:
#include < bits / stdc++.h >
using namespace std;
unsigned long int getMax(int n, int * s, int * l, int * r) {
unsigned long int dyn[n + 1] = {};
dyn[1] = s[1];
for (int i = 2; i <= n; i++) {
dyn[i] = dyn[i - 1];
unsigned long int onInc = s[i];
int left = i - l[i] - 1;
if (left >= 1) {
unsigned int k = left;
while ((k > 0) && ((r[k] + k) >= i)) {
k--;
}
if (k != 0) {
if ((dyn[k] + s[i]) > dyn[i]) {
onInc = dyn[k] + s[i];
}
}
}
dyn[i] = (dyn[i] > onInc) ? dyn[i] : onInc;
}
return dyn[n];
}
int main() {
int n;
cin >> n;
int s[n + 1] = {}, l[n + 1] = {}, r[n + 1] = {};
for (int i = 1; i <= n; i++) {
cin >> s[i];
}
for (int i = 1; i <= n; i++) {
cin >> l[i];
}
for (int i = 1; i <= n; i++) {
cin >> r[i];
}
cout << getMax(n, s, l, r) << endl;
return 0;
}
Problem in your approach:
In your DP table, the information you are storing is about maximum possible so far. The information regarding whether the ith index has been considered is lost. You can consider taking strength at current index to extend previous indices only if any of the previously seen indices is either not in range or it is in range and has not been considered.
Solution:
Reconfigure your DP recurrence. Let DP[i] denote the maximum answer if ith index was considered. Now you will only need to extend those that satisfy range condition. The answer would be maximum value of all DP indices.
Code:
vector<long> DP(n,0);
DP[0]=strength[0]; // base condition
for(int i = 1; i < n ; i++){
DP[i] = strength[i];
for(int j = 0; j < i ; j++){
if(j >= (i-l[i]) || i <= (j+r[j])){ // can't extend
}
else{
DP[i]=max(DP[i],strength[i]+DP[j]); // extend to maximize result
}
}
}
long ans=*max_element(DP.begin(),DP.end());
Time Complexity: O(n^2)
Possible Optimizations:
There are better ways to calculate maximum values which you might want to look into. You can start by looking into Segment tree and Binary Indexed Trees.