Minimum count of 0s to be removed from given Binary string to make all 1s occurs consecutively

• Difficulty Level : Medium
• Last Updated : 11 Sep, 2022

Given a binary string S of size N, the task is to find the minimum numbers of 0s that must be removed from the string S such that all the 1s occurs consecutively.

Examples:

Input: S = “010001011”
Output:
Explanation:
Removing the characters { S[2], S[3], S[4], S[6] } from the string S modifies the string S to “01111”.
Therefore, the required output is 4.

Input: S = “011110000”
Output:
Explanation:
All 1s in S already group together, therefore, the required output is 0.

Approach: The given problem can be solved by observing the fact that removing leading and ending 0s in the string doesn’t minimizes the count of 0s that must be removed. Therefore, the idea is to find the first and the last occurrence of 1 in the string S and find the count of 0s between them. Follow the steps below to solve the problem:

• Iterate over the characters of the string, S from left to right and find the index of first occurrence of 1s in the given string say X.
• Traverse the string from right to left and find the index of last occurrence of 1s in the given string say Y.
• After completing the above steps, print the count of 0s between the index X and Y as the result.

C++

 `// C++ program for the above approach``#include ``using` `namespace` `std;` `// Function to find minimum count of``// 0s removed from the string S such``// that all 1s occurs consecutively``void` `makeContiguous(string S, ``int` `N)``{``    ``// Stores the first and the last``    ``// occurrence of 1``    ``int` `fst_occur, lst_occur;` `    ``// Iterate over the characters``    ``// the string, S``    ``for` `(``int` `x = 0; x < N; x++) {` `        ``// If current character``        ``// is '1'``        ``if` `(S[x] == ``'1'``) {` `            ``// Update fst_occur``            ``fst_occur = x;``            ``break``;``        ``}``    ``}` `    ``// Iterate over the characters``    ``// the string, S``    ``for` `(``int` `x = N - 1; x >= 0; x--) {` `        ``// If current character``        ``// is '1'``        ``if` `(S[x] == ``'1'``) {` `            ``// Update lst_occur``            ``lst_occur = x;``            ``break``;``        ``}``    ``}` `    ``// Stores the count of 0s between``    ``// fst_occur and lst_occur``    ``int` `count = 0;` `    ``// Iterate over the characters of S``    ``// between fst_occur and lst_occur``    ``for` `(``int` `x = fst_occur;``        ``x <= lst_occur; x++) {` `        ``// If current character is '0'``        ``if` `(S[x] == ``'0'``) {` `            ``// Update count``            ``count++;``        ``}``    ``}` `    ``// Print the resultant minimum count``    ``cout << count;``}` `// Driver Code``int` `main()``{``    ``string S = ``"010001011"``;``    ``int` `N = S.size();``    ``makeContiguous(S, N);` `    ``return` `0;``}`

Java

 `// Java program for the above approach` `import` `java.util.*;` `class` `GFG{` `// Function to find minimum count of``// 0s removed from the String S such``// that all 1s occurs consecutively``static` `void` `makeContiguous(String S, ``int` `N)``{``    ``// Stores the first and the last``    ``// occurrence of 1``    ``int` `fst_occur=``0``, lst_occur=``0``;` `    ``// Iterate over the characters``    ``// the String, S``    ``for` `(``int` `x = ``0``; x < N; x++) {` `        ``// If current character``        ``// is '1'``        ``if` `(S.charAt(x) == ``'1'``) {` `            ``// Update fst_occur``            ``fst_occur = x;``            ``break``;``        ``}``    ``}` `    ``// Iterate over the characters``    ``// the String, S``    ``for` `(``int` `x = N - ``1``; x >= ``0``; x--) {` `        ``// If current character``        ``// is '1'``        ``if` `(S.charAt(x) == ``'1'``) {` `            ``// Update lst_occur``            ``lst_occur = x;``            ``break``;``        ``}``    ``}` `    ``// Stores the count of 0s between``    ``// fst_occur and lst_occur``    ``int` `count = ``0``;` `    ``// Iterate over the characters of S``    ``// between fst_occur and lst_occur``    ``for` `(``int` `x = fst_occur; x <= lst_occur; x++) {` `        ``// If current character is '0'``        ``if` `(S.charAt(x) == ``'0'``) {` `            ``// Update count``            ``count++;``        ``}``    ``}` `    ``// Print the resultant minimum count``    ``System.out.print(count);``}` `// Driver Code``public` `static` `void` `main(String[] args)``{``    ``String S = ``"010001011"``;``    ``int` `N = S.length();``    ``makeContiguous(S, N);` `}``}` `// This code is contributed by gauravrajput1`

Python3

 `# Python3 program for the above approach` `# Function to find minimum count of``# 0s removed from the string S such``# that all 1s occurs consecutively``def` `makeContiguous(S, N):``  ` `    ``# Stores the first and the last``    ``# occurrence of 1``    ``fst_occur ``=` `0``    ``lst_occur ``=` `0` `    ``# Iterate over the characters``    ``# the string, S``    ``for` `x ``in` `range``(N):``      ` `        ``# If current character``        ``# is '1'``        ``if` `(S[x] ``=``=` `'1'``):``          ` `            ``# Update fst_occur``            ``fst_occur ``=` `x``            ``break` `    ``# Iterate over the characters``    ``# the string, S``    ``x ``=` `N ``-` `1``    ``while``(x >``=` `0``):``      ` `        ``# If current character``        ``# is '1'``        ``if` `(S[x] ``=``=` `'1'``):``          ` `            ``# Update lst_occur``            ``lst_occur ``=` `x``            ``break``        ``x ``-``=` `1` `    ``# Stores the count of 0s between``    ``# fst_occur and lst_occur``    ``count ``=` `0` `    ``# Iterate over the characters of S``    ``# between fst_occur and lst_occur``    ``for` `x ``in` `range``(fst_occur,lst_occur``+``1``,``1``):``        ``# If current character is '0'``        ``if` `(S[x] ``=``=` `'0'``):``            ``# Update count``            ``count ``+``=` `1` `    ``# Print the resultant minimum count``    ``print``(count)` `# Driver Code``if` `__name__ ``=``=` `'__main__'``:``    ``S ``=` `"010001011"``    ``N ``=` `len``(S)``    ``makeContiguous(S, N)``    ` `    ``# This code is contributed by SURENDRA_GANGWAR.`

C#

 `// C# program for the above approach``using` `System;` `class` `GFG{` `// Function to find minimum count of``// 0s removed from the string S such``// that all 1s occurs consecutively``static` `void` `makeContiguous(``string` `S, ``int` `N)``{``    ` `    ``// Stores the first and the last``    ``// occurrence of 1``    ``int` `fst_occur = 0, lst_occur = 0;` `    ``// Iterate over the characters``    ``// the string, S``    ``for``(``int` `x = 0; x < N; x++)``    ``{``        ` `        ``// If current character``        ``// is '1'``        ``if` `(S[x] == ``'1'``)``        ``{``            ` `            ``// Update fst_occur``            ``fst_occur = x;``            ``break``;``        ``}``    ``}` `    ``// Iterate over the characters``    ``// the string, S``    ``for``(``int` `x = N - 1; x >= 0; x--)``    ``{``        ` `        ``// If current character``        ``// is '1'``        ``if` `(S[x] == ``'1'``)``        ``{``            ` `            ``// Update lst_occur``            ``lst_occur = x;``            ``break``;``        ``}``    ``}` `    ``// Stores the count of 0s between``    ``// fst_occur and lst_occur``    ``int` `count = 0;` `    ``// Iterate over the characters of S``    ``// between fst_occur and lst_occur``    ``for``(``int` `x = fst_occur; x <= lst_occur; x++)``    ``{``        ` `        ``// If current character is '0'``        ``if` `(S[x] == ``'0'``)``        ``{``            ` `            ``// Update count``            ``count++;``        ``}``    ``}` `    ``// Print the resultant minimum count``    ``Console.Write(count);``}` `// Driver Code``public` `static` `void` `Main()``{``    ``string` `S = ``"010001011"``;``    ``int` `N = S.Length;``    ` `    ``makeContiguous(S, N);``}``}` `// This code is contributed by sanjoy_62`

Javascript

 ``

Output

`4`

Time complexity: O(N) where N is the length of the given binary string
Auxiliary space: O(1), as constant extra space is required

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