Minimum swaps required to make a binary string alternating

23 Sep, 2021
You are given a binary string of even length and equal number of 0’s and 1’s. What is the minimum number of swaps to make the string alternating. A binary string is alternating if no two consecutive elements are equal.
Examples: 
Input : 000111
Output : 1
Explanation : Swap index 2 and index 5 to get 010101

Input : 1010
Output : 0 
We may either get a 1 at first position or a zero at first position. We consider two cases and find the minimum of two cases. Note that it is given that there are equal of numbers of 1’s and 0’s in the string and string is of even length.1. Count number of zeroes at odd position and even position of the string. Let their count be odd_0 and even_0 respectively. 2. Count number of ones at odd position and even position of the string. Let their count be odd_1 and even_1 respectively. 3. We will always swap a 1 with a 0 (never a 1 with a 1 or a 0 with a 0). So we just check if our alternating string starts with 0 then the number of swaps is min(even_0, odd_1) and if our alternating string starts with 1 then the number of swaps is min(even_1, odd_0). The answer is min of these two.
Below is the implementation of above approach:
C++
Java
Python3
C#
PHP
Javascript
// CPP implementation of the approach#include<bits/stdc++.h>using namespace std;     // returns the minimum number of swaps    // of a binary string    // passed as the argument    // to make it alternating    int countMinSwaps(string st)    {         int min_swaps = 0;         // counts number of zeroes at odd        // and even positions        int odd_0 = 0, even_0 = 0;         // counts number of ones at odd        // and even positions        int odd_1 = 0, even_1 = 0;         int n = st.length();        for (int i = 0; i < n; i++) {            if (i % 2 == 0) {                if (st[i] == '1')                    even_1++;                else                    even_0++;            }            else {                if (st[i] == '1')                    odd_1++;                else                    odd_0++;            }        }         // alternating string starts with 0        int cnt_swaps_1 = min(even_0, odd_1);         // alternating string starts with 1        int cnt_swaps_2 = min(even_1, odd_0);         // calculates the minimum number of swaps        return min(cnt_swaps_1, cnt_swaps_2);    }     // Driver code    int main()    {        string st = "000111";        cout<<countMinSwaps(st)<<endl;          return 0;    } // This code is contributed by Surendra_Gangwar
// Java implementation of the approach class GFG {     // returns the minimum number of swaps    // of a binary string    // passed as the argument    // to make it alternating    static int countMinSwaps(String st)    {         int min_swaps = 0;         // counts number of zeroes at odd        // and even positions        int odd_0 = 0, even_0 = 0;         // counts number of ones at odd        // and even positions        int odd_1 = 0, even_1 = 0;         int n = st.length();        for (int i = 0; i < n; i++) {            if (i % 2 == 0) {                if (st.charAt(i) == '1')                    even_1++;                else                    even_0++;            }            else {                if (st.charAt(i) == '1')                    odd_1++;                else                    odd_0++;            }        }         // alternating string starts with 0        int cnt_swaps_1 = Math.min(even_0, odd_1);         // alternating string starts with 1        int cnt_swaps_2 = Math.min(even_1, odd_0);         // calculates the minimum number of swaps        return Math.min(cnt_swaps_1, cnt_swaps_2);    }     // Driver code    public static void main(String[] args)    {        String st = "000111";        System.out.println(countMinSwaps(st));    }}
# Python3 implementation of the# above approach # returns the minimum number of swaps# of a binary string# passed as the argument# to make it alternatingdef countMinSwaps(st) :     min_swaps = 0     # counts number of zeroes at odd    # and even positions    odd_0, even_0 = 0, 0     # counts number of ones at odd    # and even positions    odd_1, even_1 = 0, 0     n = len(st)     for i in range(0, n) :         if i % 2 == 0 :             if st[i] == "1" :                even_1 += 1            else :                even_0 += 1                         else :            if st[i] == "1" :                odd_1 += 1            else :                odd_0 += 1     # alternating string starts with 0    cnt_swaps_1 = min(even_0, odd_1)     # alternating string starts with 1    cnt_swaps_2 = min(even_1, odd_0)     # calculates the minimum number of swaps    return min(cnt_swaps_1, cnt_swaps_2) # Driver code    if __name__ == "__main__" :     st = "000111"     # Function call    print(countMinSwaps(st)) # This code is contributed by# ANKITRAI1
// C# implementation of the approachusing System; public class GFG{     // returns the minimum number of swaps    // of a binary string    // passed as the argument    // to make it alternating    public static int countMinSwaps(string st)    {         int min_swaps = 0;         // counts number of zeroes at odd         // and even positions        int odd_0 = 0, even_0 = 0;         // counts number of ones at odd         // and even positions        int odd_1 = 0, even_1 = 0;         int n = st.Length;        for (int i = 0; i < n; i++)        {            if (i % 2 == 0)            {                if (st[i] == '1')                {                    even_1++;                }                else                {                    even_0++;                }            }            else            {                if (st[i] == '1')                {                    odd_1++;                }                else                {                    odd_0++;                }            }        }         // alternating string starts with 0        int cnt_swaps_1 = Math.Min(even_0, odd_1);         // alternating string starts with 1        int cnt_swaps_2 = Math.Min(even_1, odd_0);         // calculates the minimum number of swaps        return Math.Min(cnt_swaps_1, cnt_swaps_2);    }     // Driver code    public static void Main(string[] args)    {        string st = "000111";        Console.WriteLine(countMinSwaps(st));    }} // This code is contributed by Shrikant13
<?php// PHP implementation of the approach// returns the minimum number of swaps// of a binary string passed as the// argument to make it alternatingfunction countMinSwaps($st){    $min_swaps = 0;     // counts number of zeroes at odd    // and even positions    $odd_0 = 0;    $even_0 = 0;     // counts number of ones at odd    // and even positions    $odd_1 = 0;    $even_1 = 0;     $n = strlen($st);    for ($i = 0; $i < $n; $i++)    {        if ($i % 2 == 0)        {            if ($st[$i] == '1')            {                $even_1++;            }            else            {                $even_0++;            }        }        else        {            if ($st[$i] == '1')            {                $odd_1++;            }            else            {                $odd_0++;            }        }    }     // alternating string starts with 0    $cnt_swaps_1 = min($even_0, $odd_1);     // alternating string starts with 1    $cnt_swaps_2 = min($even_1, $odd_0);     // calculates the minimum number of swaps    return min($cnt_swaps_1, $cnt_swaps_2);} // Driver code$st = "000111";echo (countMinSwaps($st)); // This code is contributed by Sachin.?>
<script>    // Javascript implementation of the approach         // returns the minimum number of swaps    // of a binary string    // passed as the argument    // to make it alternating    function countMinSwaps(st)    {          let min_swaps = 0;          // counts number of zeroes at odd        // and even positions        let odd_0 = 0, even_0 = 0;          // counts number of ones at odd        // and even positions        let odd_1 = 0, even_1 = 0;          let n = st.length;        for (let i = 0; i < n; i++)        {            if (i % 2 == 0)            {                if (st[i] == '1')                {                    even_1++;                }                else                {                    even_0++;                }            }            else            {                if (st[i] == '1')                {                    odd_1++;                }                else                {                    odd_0++;                }            }        }          // alternating string starts with 0        let cnt_swaps_1 = Math.min(even_0, odd_1);          // alternating string starts with 1        let cnt_swaps_2 = Math.min(even_1, odd_0);          // calculates the minimum number of swaps        return Math.min(cnt_swaps_1, cnt_swaps_2);    }         let st = "000111";      document.write(countMinSwaps(st));         // This code is contributed by divyesh072019.</script>
1
 
Approach for odd and even length string :
Let string length be N.
In this approach , we will consider three cases :1. Answer is impossible when total number of ones > total number of zeroes + 1 or total number of zeroes > total number of ones + 1.2. String is of even length:     We will count number of ones on odd positions ( one_odd ) and number of ones on even positions ( one_even ) then answer is min ( N/2 – one_odd , N/2 – one_even ) 3. String is of odd length :    Here we consider two cases :     i) total number of ones > total number of zeroes (then we have put ones on even positions) so, answer is ( (N + 1) / 2 – number of ones at even positions).    ii) total number of zeroes > total number of ones (then we have put zeroes on even positions) so, answer is ( (N + 1) / 2 – number of zeroes at even positions ). 
C++
Java
Python3
C#
Javascript
// C++ implementation of the above approach#include <bits/stdc++.h>using namespace std; // function to count minimum swaps// required to make binary string// alternatingint countMinSwaps(string s){    int N = s.size();     // stores total number of ones    int one = 0;     // stores total number of zeroes    int zero = 0;     for (int i = 0; i < N; i++) {        if (s[i] == '1')            one++;        else            zero++;    }     // checking impossible condition    if (one > zero + 1 || zero > one + 1)        return -1;     // odd length string    if (N % 2) {        // number of even positions        int num = (N + 1) / 2;         // stores number of zeroes and        // ones at even positions        int one_even = 0, zero_even = 0;         for (int i = 0; i < N; i++) {            if (i % 2 == 0) {                if (s[i] == '1')                    one_even++;                else                    zero_even++;            }        }         if (one > zero)            return num - one_even;         else            return num - zero_even;    }     // even length string    else {        // stores number of ones at odd        // and even position respectively        int one_odd = 0, one_even = 0;         for (int i = 0; i < N; i++) {            if (s[i] == '1') {                if (i % 2)                    one_odd++;                 else                    one_even++;            }        }         return min(N / 2 - one_odd, N / 2 - one_even);    }} // Driver codeint main(){    string s = "111000";     cout << countMinSwaps(s);     return 0;}
// Java implementation of the above approachimport java.util.*; class GFG{ // function to count minimum swaps// required to make binary String// alternatingstatic int countMinSwaps(String s){    int N = s.length();     // stores total number of ones    int one = 0;     // stores total number of zeroes    int zero = 0;     for (int i = 0; i < N; i++) {        if (s.charAt(i) == '1')            one++;        else            zero++;    }     // checking impossible condition    if (one > zero + 1 || zero > one + 1)        return -1;     // odd length String    if (N % 2 == 1)    {               // number of even positions        int num = (N + 1) / 2;         // stores number of zeroes and        // ones at even positions        int one_even = 0, zero_even = 0;         for (int i = 0; i < N; i++) {            if (i % 2 == 0) {                if (s.charAt(i) == '1')                    one_even++;                else                    zero_even++;            }        }         if (one > zero)            return num - one_even;         else            return num - zero_even;    }     // even length String    else    {               // stores number of ones at odd        // and even position respectively        int one_odd = 0, one_even = 0;         for (int i = 0; i < N; i++) {            if (s.charAt(i) == '1') {                if (i % 2 == 1)                    one_odd++;                 else                    one_even++;            }        }         return Math.min(N / 2 - one_odd, N / 2 - one_even);    }} // Driver codepublic static void main(String[] args){    String s = "111000";     System.out.print(countMinSwaps(s));}} // This code is contributed by gauravrajput1
# Python implementation of the above approach# function to count minimum swaps# required to make binary string# alternatingdef countMinSwaps(s):    N = len(s)     # stores total number of ones    one = 0     # stores total number of zeroes    zero = 0     for i in range(N):        if (s[i] == '1'):            one += 1        else:            zero += 1         # checking impossible condition    if (one > zero + 1 or zero > one + 1):        return -1     # odd length string    if (N % 2):               # number of even positions        num = (N + 1) / 2         # stores number of zeroes and        # ones at even positions        one_even = 0        zero_even = 0         for i in range(N):            if (i % 2 == 0):                if (s[i] == '1'):                    one_even+=1                else:                    zero_even+=1                     if (one > zero):            return num - one_even         else:            return num - zero_even         # even length string    else:        # stores number of ones at odd        # and even position respectively        one_odd = 0        one_even = 0         for i in range(N):            if (s[i] == '1'):                if (i % 2):                    one_odd+=1                 else:                    one_even+=1                     return min(N // 2 - one_odd, N // 2 - one_even) # Driver codes = "111000"print(countMinSwaps(s)) # This code is contributed by shivanisinghss2110
// C# implementation of the above approachusing System; class GFG{ // Function to count minimum swaps// required to make binary String// alternatingstatic int countMinSwaps(string s){    int N = s.Length;     // Stores total number of ones    int one = 0;     // Stores total number of zeroes    int zero = 0;     for(int i = 0; i < N; i++)    {        if (s[i] == '1')            one++;        else            zero++;    }     // Checking impossible condition    if (one > zero + 1 || zero > one + 1)        return -1;     // Odd length String    if (N % 2 == 1)    {               // Number of even positions        int num = (N + 1) / 2;         // Stores number of zeroes and        // ones at even positions        int one_even = 0, zero_even = 0;         for(int i = 0; i < N; i++)        {            if (i % 2 == 0)            {                if (s[i] == '1')                    one_even++;                else                    zero_even++;            }        }        if (one > zero)            return num - one_even;        else            return num - zero_even;    }     // Even length String    else    {                 // Stores number of ones at odd        // and even position respectively        int one_odd = 0, one_even = 0;         for(int i = 0; i < N; i++)        {            if (s[i] == '1')            {                if (i % 2 == 1)                    one_odd++;                else                    one_even++;            }        }        return Math.Min(N / 2 - one_odd,                        N / 2 - one_even);    }} // Driver codepublic static void Main(String[] args){    string s = "111000";     Console.Write(countMinSwaps(s));}} // This code is contributed by shivanisinghss2110
<script>// JavaScript implementation of the above approach// function to count minimum swaps// required to make binary String// alternatingfunction countMinSwaps(s){    var N = s.length;     // stores total number of ones    var one = 0;     // stores total number of zeroes    var zero = 0;     for (var i = 0; i < N; i++) {        if (s.charAt(i) == '1')            one++;        else            zero++;    }     // checking impossible condition    if (one > zero + 1 || zero > one + 1)        return -1;     // odd length String    if (N % 2 == 1)    {               // number of even positions        var num = (N + 1) / 2;         // stores number of zeroes and        // ones at even positions        var one_even = 0, zero_even = 0;         for (var i = 0; i < N; i++) {            if (i % 2 == 0) {                if (s.charAt(i) == '1')                    one_even++;                else                    zero_even++;            }        }         if (one > zero)            return num - one_even;         else            return num - zero_even;    }     // even length String    else    {               // stores number of ones at odd        // and even position respectively        var one_odd = 0, one_even = 0;         for (var i = 0; i < N; i++) {            if (s.charAt(i) == '1') {                if (i % 2 == 1)                    one_odd++;                 else                    one_even++;            }        }         return Math.min(N / 2 - one_odd, N / 2 - one_even);    }} // Driver codevar s = "111000"; document.write(countMinSwaps(s)); // This code is contributed by shivanisinghss2110 </script>
1
Time Complexity : O(string length)
ankthon
shrikanth13
Sach_Code
SURENDRA_GANGWAR
pawanharwani11
divyesh072019
GauravRajput1
shivanisinghss2110
binary-string
Greedy
Strings
Strings
Greedy
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