"Minimum Operations to Make Array Equal" Problem Solving

src/main/java/com/company/minoperation/MinOperationV1.java

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+package com.company.minoperation;
+
+
+/**
+ * TimeComplexity O(N)
+ */
+public class MinOperationV1 {
+    public int minOperations(int n) {
+
+        // [1, 3, 5, 7] leftValue: 3, rightValue: 5
+        //
+        // [1, 3, 5, 7, 9] leftValue: 3, righValue: 7
+        int ret = 0;
+        int mIdx = n/2; // 2
+
+
+        int x; //왼쪽으로 이동할 예정 --  // 2
+        int y; // 오른쪽으로 이동할 예정 ++// 2
+        if (n % 2 == 1) {
+            x = mIdx - 1;
+            y = mIdx + 1;
+        } else {
+            x = mIdx - 1;
+            y = mIdx;
+        }
+
+        int leftValue = getValue(x); //
+        int rightValue = getValue(y); //
+
+        while(true) {
+
+            if(x >= 0) {
+                // Move left
+                leftValue = getValue(x--); // leftValue = 1 x = 0 -> x = -1
+            }
+
+            if (y < n) {
+                rightValue = getValue(y++); // rightValue = 5, y = 2 -> y = 3
+            }
+
+            while (true) {
+                leftValue++;  // 4
+                rightValue--; // 6
+                ret++; // ret = 1
+                if(leftValue == rightValue) {
+                    break;
+                }
+            }
+            if(x < 0) {
+                break;
+            }
+        }
+        return ret;
+    }
+
+    private int getValue(int idx) {
+        return (2 * idx) + 1;
+    }
+}

src/main/java/com/company/minoperation/MinOperationV2.java

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+package com.company.minoperation;
+
+/**
+ * Time complexity O(1)
+ */
+public class MinOperationV2 {
+
+    // [1 (2), 3,(2) 5]  answer =2
+    //[1(4), 3(2), 5, (2) 7,(4) 9]  = answer=6
+    //[1(6), 3(4), 5(2),  7, 9, 11, 13]  2 * 1 + 2* 2, ..... + 2 * (n/2)
+
+    // O(n)
+    public int minOperations(int n) {
+        if (n == 1) {
+            return 0;
+        }
+        if (n % 2 != 0) {
+            return sumOdds(n/2);
+        } else {
+            return sumEvens(n/2);
+        }
+    }
+
+    private int sumOdds(int n) {
+        return n * (n + 1);
+    }
+
+    private int sumEvens(int n) {
+        return n * (n-1 ) + n;
+    }
+
+    /*
+    홀수: 2 6 12 20 = 시그마 (2*n) -> 2 * 시그마(n) =  (n-1) * (n)
+          2 4 6 8 -> a(n) = 2*n
+    짝수: 1 4 9 16 -> 시그마 (2*n + 1) ->  1 + n * (n-1) + (n-1) = n * (n-1) + n
+          3 5 7 9 -> a(n) = 2*n + 1
+     */
+}

src/main/java/com/company/minoperation/Problem.md

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+# Minimum Operations to Make Array Equal
+
+You have an array arr of length n where arr[i] = (2 * i) + 1 for all valid values of i (i.e. 0 <= i < n).
+
+In one operation, you can select two indices x and y where 0 <= x, y < n and subtract 1 from arr[x] and add 1 to arr[y] (i.e. perform arr[x] -=1 and arr[y] += 1). The goal is to make all the elements of the array equal. It is guaranteed that all the elements of the array can be made equal using some operations.
+
+Given an integer n, the length of the array. Return the minimum number of operations needed to make all the elements of arr equal.
+
+```
+Input: n = 3
+Output: 2
+Explanation: arr = [1, 3, 5]
+First operation choose x = 2 and y = 0, this leads arr to be [2, 3, 4]
+In the second operation choose x = 2 and y = 0 again, thus arr = [3, 3, 3].
+```
+
+Source: 
+leetcode.com