Given a binary array arr[] of zero’s and one’s only. The task is to find the minimum number of one’s to be changed to zero such if there exist any index 
in the array such that arr[i] = 0 then arr[i-1] and arr[i+1] both should not be equals to 
at the same time.
That is, for any index the below condition should fail:
if (arr[i]== 0):
(arr[i-1] == 1) && (arr[i+1] == 1)
Note: 1-based indexing is considered for the array.
Examples:
Input : arr[] = { 1, 1, 0, 1, 1, 0, 1, 0, 1, 0 }
Output : 2
Explanation: Indexes 2 and 7 OR 4 and 7 can be changed to zero.Input : arr[] = { 1, 1, 0, 0, 0 }
Output : 0
Approach: The idea is that, whenever we found condition like ![arr[i-1] = 1 \;and\; arr[i] = 0 \;and \;arr[i+1] = 1](https://neveropen.co.za/wp-content/uploads/2023/11/neveropen_quicklatex.com-1c25b40ec83bb84086c857ed4ac92cf4_l3.png "Rendered by QuickLaTeX.com")
we simply changed the value of (i+1)th index to zero(0). So that index between (i-1)-th and (i+1)-th index is safe.
Below is the implementation of the above approach:
C++
// C++ program to find minimum number// of 1's to be replaced to 0's#include <bits/stdc++.h>using namespace std;// Function to find minimum number// of 1's to be replaced to 0'sint minChanges(int A[], int n){ int cnt = 0; for (int i = 0; i < n - 2; ++i) { if ((i - 1 >= 0) && A[i - 1] == 1 && A[i + 1] == 1 && A[i] == 0) { A[i + 1] = 0; cnt++; } } // return final answer return cnt;}// Driver programint main(){ int A[] = { 1, 1, 0, 1, 1, 0, 1, 0, 1, 0 }; int n = sizeof(A) / sizeof(A[0]); cout << minChanges(A, n); return 0;} |
Java
// Java program to find minimum number// of 1's to be replaced to 0'simport java.lang.*;import java.util.*;class GFG{// Function to find minimum number// of 1's to be replaced to 0'sstatic int minChanges(int[] A, int n){ int cnt = 0; for (int i = 0; i < n - 2; ++i) { if ((i - 1 >= 0) && A[i - 1] == 1 && A[i + 1] == 1 && A[i] == 0) { A[i + 1] = 0; cnt++; } } // return final answer return cnt;}// Driver Codepublic static void main(String args[]){ int[] A = { 1, 1, 0, 1, 1, 0, 1, 0, 1, 0 }; int n = A.length; System.out.print(minChanges(A, n));}}// This code is contributed// by Akanksha Rai |
Python3
# Python 3 program to find minimum # number of 1's to be replaced to 0's # Function to find minimum number# of 1's to be replaced to 0'sdef minChanges(A, n): cnt = 0 for i in range(n - 2): if ((i - 1 >= 0) and A[i - 1] == 1 and A[i + 1] == 1 and A[i] == 0): A[i + 1] = 0 cnt = cnt + 1 # return final answer return cnt # Driver CodeA = [1, 1, 0, 1, 1, 0, 1, 0, 1, 0]n = len(A)print(minChanges(A, n)) # This code is contributed # by Shashank_Sharma |
C#
// C# program to find minimum number// of 1's to be replaced to 0'susing System;class GFG{// Function to find minimum number// of 1's to be replaced to 0'sstatic int minChanges(int[] A, int n){ int cnt = 0; for (int i = 0; i < n - 2; ++i) { if ((i - 1 >= 0) && A[i - 1] == 1 && A[i + 1] == 1 && A[i] == 0) { A[i + 1] = 0; cnt++; } } // return final answer return cnt;}// Driver Codepublic static void Main(){ int[] A = { 1, 1, 0, 1, 1, 0, 1, 0, 1, 0 }; int n = A.Length; Console.Write(minChanges(A, n));}}// This code is contributed// by Akanksha Rai |
PHP
<?php// PHP program to find minimum number// of 1's to be replaced to 0's// Function to find minimum number// of 1's to be replaced to 0'sfunction minChanges($A, $n){ $cnt = 0; for ($i = 0; $i < $n - 2; ++$i) { if (($i - 1 >= 0) && $A[$i - 1] == 1 && $A[$i + 1] == 1 && $A[$i] == 0) { $A[$i + 1] = 0; $cnt++; } } // return final answer return $cnt;}// Driver Code$A = array(1, 1, 0, 1, 1, 0, 1, 0, 1, 0);$n = sizeof($A);echo minChanges($A, $n);// This code is contributed // by Ankita_Saini?> |
Javascript
<script>// Javascript program to find minimum number// of 1's to be replaced to 0's// Function to find minimum number// of 1's to be replaced to 0'sfunction minChanges(A, n){ var cnt = 0; for (var i = 0; i < n - 2; ++i) { if ((i - 1 >= 0) && A[i - 1] == 1 && A[i + 1] == 1 && A[i] == 0) { A[i + 1] = 0; cnt++; } } // return final answer return cnt;}var A = [ 1, 1, 0, 1, 1, 0, 1, 0, 1, 0 ];var n = A.length;document.write( minChanges(A, n));// This code is contributed by SoumikMondal</script> |
Output
2
Complexity Analysis:
- Time Complexity: O(N)
- Auxiliary Space: O(1)
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