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# Minimum number of flips to make a Binary String increasing

• Difficulty Level : Easy
• Last Updated : 07 Mar, 2022

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 = ‘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 ``using` `namespace` `std;` `// Function to find the minimum number of``// flips required to make string increasing``int` `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 increasing``def` `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 Code``S ``=` `'000110'` `# Function Call``print``(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

 ``

Output:

`1`

Time Complexity: O(N)
Auxiliary Space: O(1)

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