Minimum number of flips with rotation to make binary string alternating - GeeksforGeeks

30 Nov, 2021
Given a binary string S of 0s and 1s. The task is to make the given string a sequence of alternate characters by using the below operations:
Remove some prefix from the start and append it to the end.
Flip some or every bit in the given string.
Print the minimum number of bits to be flipped to make the given string alternating.
Examples:
Input: S = “001”Output: 0Explanation:No need to flip any element we can get alternating sequence by using left rotation: 010.
Input: S = “000001100”Output: 3Explanation:Following steps to find minimum flips to get alternating string:1. After rotating string 6 times towards left we will get: 100000001 2. Now we can apply flip operation as following: 101000001 -> 101010001 -> 101010101Thus, minimum flips to make string alternating is 3.
Naive Approach: The naive approach is to take all N possible combinations and calculate the minimum number of bits To flip in each of those strings. Print the minimum count among all such combinations.Time Complexity: O(N2), where N is the length of the string.Auxiliary Space: O(N)
Efficient Approach: This can be solved by observing that the final string will be either of type “101010...” or “010101...” such that all 1s will either be at odd positions or at even positions. Follow the below steps to solve the problem:
Create a prefix sum array where pref[i] means a number of changes required until index i.Create prefix arrays for both the above patterns.Check for every i, if substring[0, i] is appended at the end how many characters to be flipped required.Print the minimum number of flips among all the substrings in the above steps.
Create a prefix sum array where pref[i] means a number of changes required until index i.
Create prefix arrays for both the above patterns.
Check for every i, if substring[0, i] is appended at the end how many characters to be flipped required.
Print the minimum number of flips among all the substrings in the above steps.
Why only odd length strings are considered for rotation and why rotation will have no effect on even length string? 
I’ll try to explain this with the below example,
Suppose you have the sequence 011000 which is of even length. Without rotation, you can change it to 010101 or 101010. Now suppose you choose to append 01 at the end. The sequence becomes 100001 which can be changed to 010101 or 101010. If you compare each character, you will see that this is the same case as that of without rotation. 1000 corresponds to either 0101 or 1010 in both cases and 01 to 01 or 10.
But now consider an odd length case, 01100. Without rotation, you can change it to 01010 or 10101. Now suppose you choose to append 01 at the end. The sequence becomes 10001 which can be changed to 01010 or 10101. Now if you compare each character, you will see that 100 corresponds to 010 or 101 in both cases but 01 corresponds to 01 when 100 is 010 in case of no rotation and 101 in case of rotation.
Below is the implementation of the above approach:
C++
Java
Python3
C#
Javascript
// C++ program for the above approach#include <bits/stdc++.h>using namespace std; // Function that finds the minimum// number of flips to make the// binary string alternating if// left circular rotation is allowedint MinimumFlips(string s, int n){    int a[n];     for(int i = 0; i < n; i++)    {        a[i] = (s[i] == '1' ? 1 : 0);    }     // Initialize prefix arrays to store    // number of changes required to put    // 1s at either even or odd position    int oddone[n + 1];    int evenone[n + 1];     oddone[0] = 0;    evenone[0] = 0;     for(int i = 0; i < n; i++)    {                 // If i is odd        if (i % 2 != 0)        {                         // Update the oddone            // and evenone count            oddone[i + 1] = oddone[i] +                                (a[i] == 1 ? 1 : 0);            evenone[i + 1] = evenone[i] +                                  (a[i] == 0 ? 1 : 0);        }         // Else i is even        else        {                         // Update the oddone            // and evenone count            oddone[i + 1] = oddone[i] +                                (a[i] == 0 ? 1 : 0);            evenone[i + 1] = evenone[i] +                                  (a[i] == 1 ? 1 : 0);        }    }     // Initialize minimum flips    int minimum = min(oddone[n], evenone[n]);     // Check if substring[0, i] is    // appended at end how many    // changes will be required    for(int i = 0; i < n; i++)    {        if (n % 2 != 0)        {            minimum = min(minimum,                          oddone[n] -                          oddone[i + 1] +                         evenone[i + 1]);            minimum = min(minimum,                          evenone[n] -                          evenone[i + 1] +                           oddone[i + 1]);        }    }     // Return minimum flips    return minimum;} // Driver Codeint main(){         // Given String    string S = "000001100";     // Length of given string    int n = S.length();     // Function call    cout << (MinimumFlips(S, n));} // This code is contributed by chitranayal
// Java program for the above approachimport java.util.*; class GFG {     // Function that finds the minimum    // number of flips to make the    // binary string alternating if    // left circular rotation is allowed    static int MinimumFlips(String s, int n)    {        int[] a = new int[n];         for (int i = 0; i < n; i++) {            a[i] = (s.charAt(i) == '1' ? 1 : 0);        }         // Initialize prefix arrays to store        // number of changes required to put        // 1s at either even or odd position        int[] oddone = new int[n + 1];        int[] evenone = new int[n + 1];         oddone[0] = 0;        evenone[0] = 0;         for (int i = 0; i < n; i++) {             // If i is odd            if (i % 2 != 0) {                 // Update the oddone                // and evenone count                oddone[i + 1]                    = oddone[i]                      + (a[i] == 1 ? 1 : 0);                evenone[i + 1]                    = evenone[i]                      + (a[i] == 0 ? 1 : 0);            }             // Else i is even            else {                 // Update the oddone                // and evenone count                oddone[i + 1]                    = oddone[i]                      + (a[i] == 0 ? 1 : 0);                evenone[i + 1]                    = evenone[i]                      + (a[i] == 1 ? 1 : 0);            }        }         // Initialize minimum flips        int minimum = Math.min(oddone[n],                               evenone[n]);         // Check if substring[0, i] is        // appended at end how many        // changes will be required        for (int i = 0; i < n; i++) {            if (n % 2 != 0) {                minimum = Math.min(minimum,                                   oddone[n]                                       - oddone[i + 1]                                       + evenone[i + 1]);                minimum = Math.min(minimum,                                   evenone[n]                                       - evenone[i + 1]                                       + oddone[i + 1]);            }        }         // Return minimum flips        return minimum;    }     // Driver Code    public static void main(String[] args)    {        // Given String        String S = "000001100";         // Length of given string        int n = S.length();         // Function call        System.out.print(MinimumFlips(S, n));    }}
# Python3 program for the above approach # Function that finds the minimum# number of flips to make the# binary string alternating if# left circular rotation is alloweddef MinimumFlips(s, n):     a = [0] * n     for i in range(n):        a[i] = 1 if s[i] == '1' else 0     # Initialize prefix arrays to store    # number of changes required to put    # 1s at either even or odd position    oddone = [0] * (n + 1)    evenone = [0] * (n + 1)     for i in range(n):         # If i is odd        if(i % 2 != 0):             # Update the oddone            # and evenone count            if(a[i] == 1):                oddone[i + 1] = oddone[i] + 1            else:                oddone[i + 1] = oddone[i] + 0             if(a[i] == 0):                evenone[i + 1] = evenone[i] + 1            else:                evenone[i + 1] = evenone[i] + 0         # Else i is even        else:             # Update the oddone            # and evenone count            if (a[i] == 0):                oddone[i + 1] = oddone[i] + 1            else:                oddone[i + 1] = oddone[i] + 0             if (a[i] == 1):                evenone[i + 1] = evenone[i] + 1            else:                evenone[i + 1] = evenone[i] + 0     # Initialize minimum flips    minimum = min(oddone[n], evenone[n])     # Check if substring[0, i] is    # appended at end how many    # changes will be required    for i in range(n):        if(n % 2 != 0):            minimum = min(minimum,                          oddone[n] -                          oddone[i + 1] +                         evenone[i + 1])             minimum = min(minimum,                          evenone[n] -                          evenone[i + 1] +                           oddone[i + 1])     # Return minimum flips    return minimum # Driver Code # Given StringS = "000001100" # Length of given stringn = len(S) # Function callprint(MinimumFlips(S, n)) # This code is contributed by Shivam Singh
// C# program for the above approachusing System;class GFG{   // Function that finds the minimum  // number of flips to make the  // binary string alternating if  // left circular rotation is allowed  static int MinimumFlips(String s, int n)  {    int[] a = new int[n];     for (int i = 0; i < n; i++)    {      a[i] = (s[i] == '1' ? 1 : 0);    }     // Initialize prefix arrays to store    // number of changes required to put    // 1s at either even or odd position    int[] oddone = new int[n + 1];    int[] evenone = new int[n + 1];     oddone[0] = 0;    evenone[0] = 0;     for (int i = 0; i < n; i++)    {       // If i is odd      if (i % 2 != 0)      {         // Update the oddone        // and evenone count        oddone[i + 1] = oddone[i] +                         (a[i] == 1 ? 1 : 0);        evenone[i + 1] = evenone[i] +                          (a[i] == 0 ? 1 : 0);      }       // Else i is even      else      {         // Update the oddone        // and evenone count        oddone[i + 1] = oddone[i] +                         (a[i] == 0 ? 1 : 0);        evenone[i + 1] = evenone[i] +                          (a[i] == 1 ? 1 : 0);      }    }     // Initialize minimum flips    int minimum = Math.Min(oddone[n],                           evenone[n]);     // Check if substring[0, i] is    // appended at end how many    // changes will be required    for (int i = 0; i < n; i++)    {      if (n % 2 != 0)      {        minimum = Math.Min(minimum, oddone[n] -                                       oddone[i + 1] +                                    evenone[i + 1]);        minimum = Math.Min(minimum, evenone[n] -                                       evenone[i + 1] +                                       oddone[i + 1]);      }    }     // Return minimum flips    return minimum;  }   // Driver Code  public static void Main(String[] args)  {    // Given String    String S = "000001100";     // Length of given string    int n = S.Length;     // Function call    Console.Write(MinimumFlips(S, n));  }} // This code is contributed by Rajput-Ji
<script>// JavaScript program for the// above approach     // Function that finds the minimum    // number of flips to make the    // binary string alternating if    // left circular rotation is allowed    function MinimumFlips(s, n)    {        let a = Array.from({length: n+1}, (_, i) => 0);          for (let i = 0; i < n; i++) {            a[i] = (s[i] == '1' ? 1 : 0);        }          // Initialize prefix arrays to store        // number of changes required to put        // 1s at either even or odd position        let oddone = Array.from({length: n+1}, (_, i) => 0);        let evenone = Array.from({length: n+1}, (_, i) => 0);          oddone[0] = 0;        evenone[0] = 0;          for (let i = 0; i < n; i++) {              // If i is odd            if (i % 2 != 0) {                  // Update the oddone                // and evenone count                oddone[i + 1]                    = oddone[i]                      + (a[i] == 1 ? 1 : 0);                evenone[i + 1]                    = evenone[i]                      + (a[i] == 0 ? 1 : 0);            }              // Else i is even            else {                  // Update the oddone                // and evenone count                oddone[i + 1]                    = oddone[i]                      + (a[i] == 0 ? 1 : 0);                evenone[i + 1]                    = evenone[i]                      + (a[i] == 1 ? 1 : 0);            }        }          // Initialize minimum flips        let minimum = Math.min(oddone[n],                               evenone[n]);          // Check if substring[0, i] is        // appended at end how many        // changes will be required        for (let i = 0; i < n; i++) {            if (n % 2 != 0) {                minimum = Math.min(minimum,                                   oddone[n]                                       - oddone[i + 1]                                       + evenone[i + 1]);                minimum = Math.min(minimum,                                   evenone[n]                                       - evenone[i + 1]                                       + oddone[i + 1]);            }        }          // Return minimum flips        return minimum;    } // Driver Code      // Given String        let S = "000001100";          // Length of given string        let n = S.length;          // Function call        document.write(MinimumFlips(S, n)); </script>
3
 
Time Complexity: O(N), where N is the length of the given string.Auxiliary Space: O(N)
SHIVAMSINGH67
ukasp
Rajput-Ji
susmitakundugoaldanga
sahil9420
binary-string
Google
Greedy
Mathematical
Strings
Google
Strings
Greedy
Mathematical
Writing code in comment?
Please use ide.geeksforgeeks.org,
generate link and share the link here.
Huffman Coding | Greedy Algo-3
Coin Change | DP-7
Activity Selection Problem | Greedy Algo-1
Fractional Knapsack Problem
Program for Shortest Job First (or SJF) CPU Scheduling | Set 1 (Non- preemptive)
Program for Fibonacci numbers
C++ Data Types
Set in C++ Standard Template Library (STL)
Coin Change | DP-7
Merge two sorted arrays