Given a binary string S, the task is to find the minimum number of characters the needed to be flipped to make the given binary string increasing.
Example:
Input: S = “00110”
Output: 1
Explanation: Flip S[4] = ‘0’ to ‘1’ of the string modifies the given string to “00111”. Therefore, the minimum number of flips required is 1.Input: S = “010110”
Output: 2
Approach: The given problem can be solved by using a Greedy Algorithm based on the observations that the resultant monotonically increasing string after any number of flips will be of the form (‘0’*p + ‘1’*q), where p and q are the count of 0s and 1s respectively in the modified string. The idea is to traverse the given string S and for each index, i modify the substring S[0, i) to 0s and substring S[i, N) to 1s and find the minimum flips required accordingly. Follow the steps below to solve the problem:
- Find the count of 0s in the given binary string S and store it in a variable countZero.
- Initialize variable, say minFlips as (N – cntZero) that stores the minimum number of flips required.
- Initialize variable, say cntOne as 0 that stores the count of 1s in the string while traversing the string.
- Traverse the given string S and perform the following steps:
- If the character S[i] is 0, then decrement the value of countZero by 1.
- Otherwise, update the value of minFlips as the minimum of minFlips and (countZero + countOne) and increment the value of countOne by 1.
- After completing the above steps, print the value of minFlips as the result.
Below is the implementation of the above approach.
C++
// C++ program for the above approach;#include <bits/stdc++.h>using namespace std;// Function to find the minimum number of// flips required to make string increasingint minimumFlips(string s){ // Length of s int n = s.size(); // Total number of zero in s int cnt0 = count(s.begin(), s.end(), '0'); // Stores count of 1s till ith index int cnt1 = 0; // stores the minimum count of flip int res = n - cnt0; // Traverse the given string for (int i = 0; i < n; i++) { if (s[i] == '0') { cnt0 -= 1; } // Update the value of res // and count of 1s else if (s[i] == '1') { res = min(res, cnt1 + cnt0); cnt1++; } } // Return the minimum number // of flips return res;}// Driver code int main(){ // Given string string S = "000110"; // function call cout << minimumFlips(S); return 0;}// This code is contributed by parthmanchanda81 |
Java
// Java program for the above approach;import java.util.*;class GFG{ // Function to find the minimum number of // flips required to make String increasing static int minimumFlips(String s) { // Length of s int n = s.length(); // Total number of zero in s int cnt0 = count(s, '0'); // Stores count of 1s till ith index int cnt1 = 0; // stores the minimum count of flip int res = n - cnt0; // Traverse the given String for (int i = 0; i < n; i++) { if (s.charAt(i) == '0') { cnt0 -= 1; } // Update the value of res // and count of 1s else if (s.charAt(i) == '1') { res = Math.min(res, cnt1 + cnt0); cnt1++; } } // Return the minimum number // of flips return res; } private static int count(String s, char c) { int ans = 0; for (char i : s.toCharArray()) if (c == i) ans++; return ans; } // Driver code public static void main(String[] args) { // Given String String S = "000110"; // function call System.out.print(minimumFlips(S)); }}// This code is contributed by Princi Singh |
Python3
# Python program for the above approach# Function to find the minimum number of# flips required to make string increasingdef minimumFlips(s): # Length of s n = len(s) # Total number of zero in s cnt0 = s.count('0') # Stores count of 1s till ith index cnt1 = 0 # Stores the minimum count of flips res = n - cnt0 # Traverse the given string S for i in range(n): if s[i] == '0': cnt0 -= 1 elif s[i] == '1': # Update the value of res # and count of 1s res = min(res, cnt1 + cnt0) cnt1 += 1 # Return the minimum number # of flips return res# Driver CodeS = '000110'# Function Callprint(minimumFlips(S)) |
C#
using System;public class GFG { // Function to find the minimum number of // flips required to make String increasing static int minimumFlips(String s) { // Length of s int n = s.Length; // Total number of zero in s int cnt0 = count(s, '0'); // Stores count of 1s till ith index int cnt1 = 0; // stores the minimum count of flip int res = n - cnt0; // Traverse the given String for (int i = 0; i < n; i++) { if (s[i] == '0') { cnt0 -= 1; } // Update the value of res // and count of 1s else if (s[i] == '1') { res = Math.Min(res, cnt1 + cnt0); cnt1++; } } // Return the minimum number // of flips return res; } private static int count(String s, char c) { int ans = 0; for (int j = 0; j < s.Length; j++) { char i = s[j]; if (c == i) ans++; } return ans; } // Driver code static public void Main() { // Given String String S = "000110"; // function call Console.Write(minimumFlips(S)); }}// This code is contributed by maddler. |
Javascript
<script> // JavaScript program for the above approach; // Function to find the minimum number of // flips required to make string increasing function minimumFlips(s) { // Length of s let n = s.length; // Total number of zero in s let i; let cnt0 = 0; for (i = 0; i < n; i++) { if (s[i] == '0') cnt0++; } // Stores count of 1s till ith index let cnt1 = 0 // Stores the minimum count of flips let res = n - cnt0 // Traverse the given string S for (i = 0; i < n; i++) if (s[i] == '0') { cnt0 -= 1 } else if (s[i] == '1') { // Update the value of res // and count of 1s res = Math.min(res, cnt1 + cnt0) cnt1 += 1 } // Return the minimum number // of flips return res } // Driver Code S = '000110' // Function Call document.write(minimumFlips(S))// This code is contributed by Potta Lokesh </script> |
Output:
1
Time Complexity: O(N)
Auxiliary Space: O(1)
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