Given binary string str, the task is to find the minimum number of flips required to keep all 1s together in the given binary string, i.e. there must not be any 0 between 1s in the string.
Examples:
Input: str = “0011111100”
Output: 0
Explanation: We dont need to flip any bits because all the ones are grouped together and there is no zero between any two ones.Input: str = “11100111000101”
Output: 4
Explanation: We can flip the 4th and 5th bit to make them 1 and flip 12th and 14th bit to make them 0. So the resulting string is “11111111000000” with 4 possible flips.
Approach: To solve the problem mentioned above we will implement the dynamic programming approach where we will have the following states:
- The first state is dp[i][0] which signifies the number of flips required to make all zeroes up to the ith bit.
- Second state dp[i][1] which signifies the number of flips required to make the current bit 1 such that the conditions given in the question are satisfied.
So the required answer will be minimum flips for making the current bit 1 + minimum flips for making all bits after the current bit 0 for all values of i. But if all the bits in the given string are 0 then we don’t have to change anything, so we can check the minimum between our answer and the number of flips required to make the string with all zeroes. So we can compute the answer by iterating over all the characters in the string where,
answer = min ( answer, dp[i][1] + dp[n-1][0] – dp[i][0])
where
dp[i][1] = Minimum number of flips to set current bit to 1
dp[n-1][0] – dp[i][0] = Minimum number of flips required to make all bits after i as 0
Below is the implementation of the above approach:
C++
//cpp implementation for Minimum number of//flips required in a binary string such//that all the 1’s are together#include <bits/stdc++.h>using namespace std;int minFlip(string a){ //Length of the binary string int n = a.size(); vector<vector<int>> dp(n + 1,vector<int>(2, 0)); //Initial state of the dp //dp[0][0] will be 1 if the current //bit is 1 and we have to flip it dp[0][0] = (a[0] == '1'); //Initial state of the dp //dp[0][1] will be 1 if the current //bit is 0 and we have to flip it dp[0][1] = (a[0] == '0'); for (int i = 1; i < n; i++) { //dp[i][0] = Flips required to //make all previous bits zero //+ Flip required to make current bit zero dp[i][0] = dp[i - 1][0] + (a[i] == '1'); //dp[i][1] = minimum flips required //to make all previous states 0 or make //previous states 1 satisfying the condition dp[i][1] = min(dp[i - 1][0], dp[i - 1][1]) + (a[i] == '0'); } int answer = INT_MAX; for (int i=0;i<n;i++) { answer = min(answer, dp[i][1] + dp[n - 1][0] - dp[i][0]); } //Minimum of answer and flips //required to make all bits 0 return min(answer, dp[n - 1][0]);}//Driver Codeint main() { string s = "1100111000101"; cout<<(minFlip(s));}// This code is contributed by Mohit kumar 29 |
Java
// Java implementation for // Minimum number of flips // required in a binary string // such that all the 1’s are togetherimport java.io.*;import java.util.*;class GFG{ static int minFlip(String a){ // Length of the binary string int n = a.length(); int dp[][] = new int[n + 1][2]; // Initial state of the dp // dp[0][0] will be 1 if // the current bit is 1 // and we have to flip it if(a.charAt(0) == '1') { dp[0][0] = 1 ; } else dp[0][0] = 0; // Initial state of the dp // dp[0][1] will be 1 if // the current bit is 0 // and we have to flip it if(a.charAt(0) == '0') dp[0][1] = 1; else dp[0][1] = 0; for (int i = 1; i < n; i++) { // dp[i][0] = Flips required to // make all previous bits zero // Flip required to make current // bit zero if(a.charAt(i) == '1') { dp[i][0] = dp[i - 1][0] + 1; } else dp[i][0] = dp[i - 1][0]; // dp[i][1] = minimum flips // required to make all previous // states 0 or make previous states // 1 satisfying the condition if(a.charAt(i) == '0') { dp[i][1] = Math.min(dp[i - 1][0], dp[i - 1][1]) + 1; } else dp[i][1] = Math.min(dp[i - 1][0], dp[i - 1][1]); } int answer = Integer.MAX_VALUE; for (int i = 0; i < n; i++) { answer = Math.min(answer, dp[i][1] + dp[n - 1][0] - dp[i][0]); } //Minimum of answer and flips //required to make all bits 0 return Math.min(answer, dp[n - 1][0]);} // Driver codepublic static void main (String[] args) { String s = "1100111000101"; System.out.println(minFlip(s));}}// This code is contributed by satvikshrivas26 |
Python3
# Python implementation for Minimum number of# flips required in a binary string such# that all the 1’s are togetherdef minFlip(a): # Length of the binary string n = len(a) dp =[[0, 0] for i in range(n)] # Initial state of the dp # dp[0][0] will be 1 if the current # bit is 1 and we have to flip it dp[0][0]= int(a[0]=='1') # Initial state of the dp # dp[0][1] will be 1 if the current # bit is 0 and we have to flip it dp[0][1]= int(a[0]=='0') for i in range(1, n): # dp[i][0] = Flips required to # make all previous bits zero # + Flip required to make current bit zero dp[i][0]= dp[i-1][0]+int(a[i]=='1') # dp[i][1] = minimum flips required # to make all previous states 0 or make # previous states 1 satisfying the condition dp[i][1]= min(dp[i-1])+int(a[i]=='0') answer = 10**18 for i in range(n): answer = min(answer, dp[i][1]+dp[n-1][0]-dp[i][0]) # Minimum of answer and flips # required to make all bits 0 return min(answer, dp[n-1][0]) # Driver codes = "1100111000101"print(minFlip(s)) |
C#
// C# implementation for // Minimum number of // flips required in // a binary string such// that all the 1’s are togetherusing System;class GFG{ static int minFlip(string a){ //Length of the binary string int n = a.Length; int [,]dp=new int[n + 1, 2]; //Initial state of the dp //dp[0][0] will be 1 if the current //bit is 1 and we have to flip it dp[0, 0] = (a[0] == '1' ? 1 : 0); //Initial state of the dp //dp[0][1] will be 1 if the current //bit is 0 and we have to flip it dp[0, 1] = (a[0] == '0' ? 1 : 0); for (int i = 1; i < n; i++) { //dp[i][0] = Flips required to //make all previous bits zero //+ Flip required to make current bit zero dp[i, 0] = dp[i - 1, 0] + (a[i] == '1' ? 1 : 0); //dp[i][1] = minimum flips required //to make all previous states 0 or make //previous states 1 satisfying the condition dp[i, 1] = Math.Min(dp[i - 1, 0], dp[i - 1, 1]) + (a[i] == '0' ? 1 : 0); } int answer = int.MaxValue; for (int i = 0; i < n; i++) { answer = Math.Min(answer, dp[i, 1] + dp[n - 1, 0] - dp[i, 0]); } //Minimum of answer and flips //required to make all bits 0 return Math.Min(answer, dp[n - 1, 0]);} // Driver codepublic static void Main(string[] args){ string s = "1100111000101"; Console.Write(minFlip(s));}}// This code is contributed by Rutvik_56 |
Javascript
<script>// Javascript implementation for// Minimum number of flips// required in a binary string// such that all the 1’s are togetherfunction minFlip(a){ // Length of the binary string let n = a.length; let dp = new Array(n + 2).fill(0).map((t) => new Array(2).fill(0)) // Initial state of the dp // dp[0][0] will be 1 if // the current bit is 1 // and we have to flip it if (a.charAt(0) == '1') { dp[0][0] = 1; } else dp[0][0] = 0; // Initial state of the dp // dp[0][1] will be 1 if // the current bit is 0 // and we have to flip it if (a.charAt(0) == '0') dp[0][1] = 1; else dp[0][1] = 0; for (let i = 1; i < n; i++) { // dp[i][0] = Flips required to // make all previous bits zero // Flip required to make current // bit zero if (a.charAt(i) == '1') { dp[i][0] = dp[i - 1][0] + 1; } else dp[i][0] = dp[i - 1][0]; // dp[i][1] = minimum flips // required to make all previous // states 0 or make previous states // 1 satisfying the condition if (a.charAt(i) == '0') { dp[i][1] = Math.min(dp[i - 1][0], dp[i - 1][1]) + 1; } else dp[i][1] = Math.min(dp[i - 1][0], dp[i - 1][1]); } let answer = Number.MAX_SAFE_INTEGER; for (let i = 0; i < n; i++) { answer = Math.min(answer, dp[i][1] + dp[n - 1][0] - dp[i][0]); } // Minimum of answer and flips // required to make all bits 0 return Math.min(answer, dp[n - 1][0]);}// Driver codelet s = "1100111000101";document.write(minFlip(s));// This code is contributed by _saurabh_jaiswal.</script> |
Output:
4
Time Complexity: O(N)
Auxiliary Space: O(N), where N is the length of the string.
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