1 Count Problem

#1

I tried implementing this question but I could only write the code to print the length of the longest sub array with 1s.
How am I supposed to print the new array after flipping k 0s?
LINK : https://ide.codingblocks.com/s/358552

#2

hello @gptradhika1208

just use two variables one for storing the starting point of the sequence from where maxlen starts and one for storing the maximum length .

now use those variable to perform updation from 0 to 1.

this might help->

#include <iostream>
using namespace std;

void findZeros(int *arr, int n, int k) {
    int wL = 0, wR = 0, bestWindow = 0, zeroCount = 0, bestL = 0;
    while(wR < n) {
        if(zeroCount<=k) {
            if(arr[wR] == 0) 
                zeroCount++;
            wR++;
        } 
        if(zeroCount>k) {
            if(arr[wL] == 0) 
                zeroCount--;
            wL++;
        }
        if((wR-wL) > bestWindow and (zeroCount<=k)) {
            bestWindow = wR-wL;
            bestL = wL;
        }
    }
    // cout<<bestWindow<<","<<wR<<","<<wL<<","<<bestL<<endl;
    cout<<bestWindow<<endl;
    for(int i=0;i<bestL;i++) {
        cout<<arr[i]<<" ";
    }
    for(int i=bestL;i<bestL+bestWindow;i++) {
        if(arr[i]==0)
            arr[i]=1;
        cout<<arr[i]<<" ";
    }
    for(int i=bestL+bestWindow;i<n;i++) {
        cout<<arr[i]<<" ";
    }
}

int main(int argc, char const *argv[])
{
    /* code */
    int n, k;
    cin>>n>>k;
    int *arr = new int[n];
    for(int i=0;i<n;i++) {
        cin>>arr[i];
    }

    findZeros(arr, n, k);
    return 0;
}
#3

I couldn’t really undertand the code. Can you please make changes in my code itself?

#4

@gptradhika1208
refer this explanation ->
The idea is to use Sliding Window for the given array. Let us use a window covering from index wL to index wR. Let the number of zeros inside the window be zeroCount. We maintain the window with at most m zeros inside.

The main steps are:

  • While zeroCount is no more than m: expand the window to the right (wR++) and update the count zeroCount.
  • While zeroCount exceeds m, shrink the window from left (wL++), update zeroCount;
  • Update the widest window along the way. The positions of output zeros are inside the best window.

if still the code does not make sense then ping me back.

#5

I tried to modify the code a bit but I’m still not getting the required output. The final value of pointer ‘r’ is coming out to be 0 and the length is also coming out to be incorrect. Please help me correct the errors. LINK : https://ide.codingblocks.com/s/358652

#6

Please reply as soon as possible.

#7

check now->

#8

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