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: 1
Explanation: 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: 0
Explanation: 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 number
- Find the contiguous region of length m that has the most 0s in it
- The 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 togetherint 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 codeint 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 approachimport 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 togetherdef 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 codeif __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 approachusing 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> |
Output
1
Time Complexity: O(N)
Auxiliary Space: O(1)
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