Minimum swaps to group all 0s together in Binary Circular Array Last Updated : 28 Jan, 2022 Comments Improve Suggest changes Like Article Like Report Given a binary circular array arr[] of size N, the task is to find the minimum swaps to group all 0s together in the array. Examples: Input: arr[] = {1, 0, 1, 0, 0, 1, 1}Output: 1Explanation: Here are a few of the ways to group all the 0's together: {1, 1, 0, 0, 0, 1, 1} using 1 swap.{1, 0, 0, 0, 1, 1, 1} using 1 swap.{0, 0, 1, 1, 1, 1, 0} using 2 swaps (using the circular property of the array). There is no way to group all 0's together with 0 swaps. Thus, the minimum number of swaps required is 1. Input: arr[] = {0, 0, 1, 1, 0}Output: 0Explanation: All the 0's are already grouped together due to the circular property of the array. Thus, the minimum number of swaps required is 0. Approach: The task can be solved using the sliding window technique. Follow the below steps to solve the problem: Count the total number of 0s. Let m be that numberFind the contiguous region of length m that has the most 0s in itThe number of 1s in this region is the minimum required swaps. Each swap will move one 0 into the region and one 1 out of the region.Finally, use modulo operation for handling the case of the circular arrays. Below is the implementation of the above approach. C++ // C++ program for the above approach #include <bits/stdc++.h> using namespace std; // Function to count min swap // to get all zeros together int minSwaps(int nums[], int N) { int count_1 = 0; if (N == 1) return 0; for (int i = 0; i < N; i++) { if (nums[i] == 0) count_1++; } // Window size for counting // maximum no. of 1s int windowsize = count_1; count_1 = 0; for (int i = 0; i < windowsize; i++) { if (nums[i] == 0) count_1++; } // For storing maximum count of 1s in // a window int mx = count_1; for (int i = windowsize; i < N + windowsize; i++) { if (nums[i % N] == 0) count_1++; if (nums[(i - windowsize) % N] == 0) count_1--; mx = max(count_1, mx); } return windowsize - mx; } // Driver code int main() { int nums[] = { 1, 0, 1, 0, 0, 1, 1 }; int N = sizeof(nums) / sizeof(nums[0]); cout << minSwaps(nums, N); return 0; } Java // Java program for the above approach import java.io.*; import java.lang.*; import java.util.*; class GFG { // Function to count min swap // to get all zeros together static int minSwaps(int nums[], int N) { int count_1 = 0; if (N == 1) return 0; for (int i = 0; i < N; i++) { if (nums[i] == 0) count_1++; } // Window size for counting // maximum no. of 1s int windowsize = count_1; count_1 = 0; for (int i = 0; i < windowsize; i++) { if (nums[i] == 0) count_1++; } // For storing maximum count of 1s in // a window int mx = count_1; for (int i = windowsize; i < N + windowsize; i++) { if (nums[i % N] == 0) count_1++; if (nums[(i - windowsize) % N] == 0) count_1--; mx = Math.max(count_1, mx); } return windowsize - mx; } // Driver code public static void main (String[] args) { int nums[] = { 1, 0, 1, 0, 0, 1, 1 }; int N = nums.length; System.out.println( minSwaps(nums, N)); } } // This code is contributed by hrithikgarg03188. Python3 # python3 program for the above approach # Function to count min swap # to get all zeros together def minSwaps(nums, N): count_1 = 0 if (N == 1): return 0 for i in range(0, N): if (nums[i] == 0): count_1 += 1 # Window size for counting # maximum no. of 1s windowsize = count_1 count_1 = 0 for i in range(0, windowsize): if (nums[i] == 0): count_1 += 1 # For storing maximum count of 1s in # a window mx = count_1 for i in range(windowsize, N + windowsize): if (nums[i % N] == 0): count_1 += 1 if (nums[(i - windowsize) % N] == 0): count_1 -= 1 mx = max(count_1, mx) return windowsize - mx # Driver code if __name__ == "__main__": nums = [1, 0, 1, 0, 0, 1, 1] N = len(nums) print(minSwaps(nums, N)) # This code is contributed by rakeshsahni C# // C# program for the above approach using System; class GFG { // Function to count min swap // to get all zeros together static int minSwaps(int []nums, int N) { int count_1 = 0; if (N == 1) return 0; for (int i = 0; i < N; i++) { if (nums[i] == 0) count_1++; } // Window size for counting // maximum no. of 1s int windowsize = count_1; count_1 = 0; for (int i = 0; i < windowsize; i++) { if (nums[i] == 0) count_1++; } // For storing maximum count of 1s in // a window int mx = count_1; for (int i = windowsize; i < N + windowsize; i++) { if (nums[i % N] == 0) count_1++; if (nums[(i - windowsize) % N] == 0) count_1--; mx = Math.Max(count_1, mx); } return windowsize - mx; } // Driver code public static void Main() { int []nums = { 1, 0, 1, 0, 0, 1, 1 }; int N = nums.Length; Console.Write(minSwaps(nums, N)); } } // This code is contributed by Samim Hossain Mondal. JavaScript <script> // JavaScript code for the above approach // Function to count min swap // to get all zeros together function minSwaps(nums, N) { let count_1 = 0; if (N == 1) return 0; for (let i = 0; i < N; i++) { if (nums[i] == 0) count_1++; } // Window size for counting // maximum no. of 1s let windowsize = count_1; count_1 = 0; for (let i = 0; i < windowsize; i++) { if (nums[i] == 0) count_1++; } // For storing maximum count of 1s in // a window let mx = count_1; for (let i = windowsize; i < N + windowsize; i++) { if (nums[i % N] == 0) count_1++; if (nums[(i - windowsize) % N] == 0) count_1--; mx = Math.max(count_1, mx); } return windowsize - mx; } // Driver code let nums = [1, 0, 1, 0, 0, 1, 1]; let N = nums.length; document.write(minSwaps(nums, N)); // This code is contributed by Potta Lokesh </script> Output1 Time Complexity: O(N)Auxiliary Space: O(1) Comment More infoAdvertise with us Next Article Minimum swaps to group all 0s together in Binary Circular Array sauarbhyadav Follow Improve Article Tags : Mathematical Competitive Programming Algo Geek DSA Arrays binary-representation sliding-window circular-array Swap-Program +5 More Practice Tags : ArraysMathematicalsliding-window Similar Reads Maximum consecutive one’s (or zeros) in a binary circular array Given a binary circular array of size N, the task is to find the count maximum number of consecutive 1’s present in the circular array.Examples: Input: arr[] = {1, 1, 0, 0, 1, 0, 1, 0, 1, 1, 1, 1} Output: 6 The last 4 and first 2 positions have 6 consecutive ones. 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