solve prod of array except self

Author: samcho0608

product-of-array-except-self/samcho0608.java

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+// link: https://leetcode.com/problems/product-of-array-except-self/submissions/1831301674/
+// difficulty: Medium
+class Solution1 {
+    // Problem:
+    // * return: array where answer[i] = product of all elements except nums[i]
+    // Solution:
+    // * Time Complexity: O(N)
+    // * Space Complexity: O(N)
+    public int[] productExceptSelf(int[] nums) {
+        int n = nums.length;
+
+        // prefixProds[i] = prod of all elements upto i (exclusive)
+        int[] prefixProds = new int[n];
+        for(int i = 0; i < n; i++) {
+            if(i == 0) {
+                prefixProds[i] = 1;
+                continue;
+            }
+
+            prefixProds[i] = prefixProds[i-1] * nums[i-1];
+        }
+
+        // suffixProds[i] = prod of all elements from end to i (exclusive)
+        int[] suffixProds = new int[n];
+        for(int i = n - 1; i >= 0; i--) {
+            if(i == n - 1) {
+                suffixProds[i] = 1;
+                continue;
+            }
+
+            suffixProds[i] = suffixProds[i+1] * nums[i+1];
+        }
+
+        // multiply prefix and suffix prods to find answers
+        int[] answers = new int[n];
+        for(int i = 0; i< n;i++) {
+            answers[i] = prefixProds[i] * suffixProds[i];
+        }
+
+        return answers;
+    }
+}
+
+// uses only 1 array whereas solution 1 uses 3
+class Solution2 {
+    // Problem:
+    // * return: array where answer[i] = product of all elements except nums[i]
+    // Solution:
+    // * Time Complexity: O(N)
+    // * Space Complexity: O(N)
+    public int[] productExceptSelf(int[] nums) {
+        int n = nums.length;
+
+        // prefixProds[i] = prod of all elements upto i (exclusive)
+        int[] answers = new int[n];
+        for(int i = 0; i < n; i++) {
+            if(i == 0) {
+                answers[i] = 1;
+                continue;
+            }
+
+            answers[i] = answers[i-1] * nums[i-1];
+        }
+
+        // use a single variable for suffix product
+        int suffixProd = 1;
+        for(int i = n - 2; i >= 0; i--) {
+            suffixProd *= nums[i+1];
+            answers[i] *= suffixProd;
+        }
+
+        return answers;
+    }
+}
+