Couples Holding Hands

Author: gouthampradhan

README.md

@@ -230,6 +230,7 @@ My accepted leetcode solutions to some of the common interview problems.
 - [Bulb Switcher II](problems/src/math/BulbSwitcherII.java) (Medium)
 - [Global and Local Inversions](problems/src/math/GlobalAndLocalInversions.java) (Medium)
 - [Solve the Equation](problems/src/math/SolveTheEquation.java) (Medium)
+- [Couples Holding Hands](problems/src/math/CouplesHoldingHands.java) (Hard)
 
 #### [Reservoir Sampling](problems/src/reservoir_sampling)
 

problems/src/math/CouplesHoldingHands.java

@@ -0,0 +1,81 @@
+package math;
+
+/**
+ * Created by gouthamvidyapradhan on 23/06/2018.
+ * N couples sit in 2N seats arranged in a row and want to hold hands. We want to know the minimum number of swaps so
+ * that every couple is sitting side by side. A swap consists of choosing any two people, then they stand up and
+ * switch  seats.
+
+ The people and seats are represented by an integer from 0 to 2N-1, the couples are numbered in order, the first
+ couple  being (0, 1), the second couple being (2, 3), and so on with the last couple being (2N-2, 2N-1).
+
+ The couples' initial seating is given by row[i] being the value of the person who is initially sitting in the i-th
+ seat.
+
+ Example 1:
+
+ Input: row = [0, 2, 1, 3]
+ Output: 1
+ Explanation: We only need to swap the second (row[1]) and third (row[2]) person.
+ Example 2:
+
+ Input: row = [3, 2, 0, 1]
+ Output: 0
+ Explanation: All couples are already seated side by side.
+ Note:
+
+ len(row) is even and in the range of [4, 60].
+ row is guaranteed to be a permutation of 0...len(row)-1.
+
+ Solution: O(N ^ 2). Find the index i of every even-number n and (n + 1)th number. If the index i of number n is even
+ then swap the number (n + 1) with index i + 1, else swap the number (n + 1) with index i - 1. Count the total swaps
+ and return the answer.
+ */
+public class CouplesHoldingHands {
+    /**
+     * Main method
+     * @param args
+     * @throws Exception
+     */
+    public static void main(String[] args) throws Exception{
+        int[] A = {1, 3, 4, 0, 2, 5};
+        System.out.println(new CouplesHoldingHands().minSwapsCouples(A));
+    }
+
+    public int minSwapsCouples(int[] row) {
+        int N = row.length;
+        int count = 0;
+        for(int i = 0; i < N; i +=2){
+            int pos = find(row, i);
+            if((pos % 2) == 0){
+                if(row[pos + 1] != i + 1){
+                    int nexNumPos = find(row, i + 1);
+                    swap(row, pos + 1, nexNumPos);
+                    count ++;
+                }
+            } else{
+                if(row[pos - 1] != i + 1){
+                    int nexNumPos = find(row, i + 1);
+                    swap(row, pos - 1, nexNumPos);
+                    count ++;
+                }
+            }
+        }
+        return count;
+    }
+
+    private int find(int[] A, int n){
+        for(int i = 0; i < A.length; i ++){
+            if(A[i] == n){
+                return i;
+            }
+        }
+        return -1;
+    }
+
+    private void swap(int[] A, int i, int j){
+        int temp = A[i];
+        A[i] = A[j];
+        A[j] = temp;
+    }
+}