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++
// 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
// 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)); }} |
Python 3
# Python3 implementation of the # above approach # returns the minimum number of swaps # of a binary string # passed as the argument # to make it alternating def 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#
// 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// PHP 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) { $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.?> |
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