feat: solve product of array except self

Author: GangBean

product-of-array-except-self/GangBean.java

@@ -0,0 +1,56 @@
+class Solution {
+    public int[] productExceptSelf(int[] nums) {
+        /**
+            1. understanding
+            - given integer array nums
+            - product of which's all elements except that element
+            - should be under O(N) time complexity
+            - should not use division operation
+            2. strategy
+            - brute force: O(n^2)
+                - for every elements, calculate product of other elements
+                    - 1: 2 * [3 * 4]
+                    - 2: 1 * [3 * 4]
+                    - 3: [1 * 2] * 4
+                    - 4: [1 * 2] * 3
+            - dynamic programming
+                - assign array mem to memorize product till that idx
+                - mem memorize in ascending order product value and reverse order product value
+            3. complexity
+            - time: O(N)
+            - space: O(N)
+        */
+        // 1. assign array variable mem
+        int[][] mem = new int[nums.length][];
+        for (int i = 0 ; i < nums.length; i++) {
+            mem[i] = new int[2];
+        }
+
+        // 2. calculate product values
+        for (int i = 0 ; i < nums.length; i++) { // O(N)
+            if (i == 0) {
+                mem[i][0] = nums[i];
+                continue;
+            }
+            mem[i][0] = nums[i] * mem[i-1][0];
+        }
+
+        for (int i = nums.length - 1; i >= 0; i--) { // O(N)
+            if (i == nums.length - 1) {
+                mem[i][1] = nums[i];
+                continue;
+            }
+            mem[i][1] = nums[i] * mem[i+1][1];
+        }
+
+        int[] ret = new int[nums.length];
+        for (int i = 0; i < nums.length; i++) { // O(N)
+            int left = (i - 1) >= 0 ? mem[i-1][0] : 1;
+            int right = (i + 1) < nums.length ? mem[i+1][1] : 1;
+            ret[i] = left * right;
+        }
+
+        return ret;
+    }
+}
+