Minimum count of 0s to be removed from given Binary string to make all 1s occurs consecutively
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: 4
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: 0
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 <bits/stdc++.h>
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 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 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# 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
<script>
// javascript program for the above approach
// Function to find minimum count of
// 0s removed from the String S such
// that all 1s occurs consecutively
function makeContiguous( S , N)
{
// Stores the first and the last
// occurrence of 1
var fst_occur = 0, lst_occur = 0;
// Iterate over the characters
// the String, S
for (var 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 (var 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
var count = 0;
// Iterate over the characters of S
// between fst_occur and lst_occur
for (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
document.write(count);
}
// Driver Code
var S = "010001011";
var N = S.length;
makeContiguous(S, N);
// This code is contributed by umadevi9616
</script>
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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