[LeetCode] Find Pivot Index

Given an array of integers nums, write a method that returns the "pivot" index of this array.

We define the pivot index as the index where the sum of the numbers to the left of the index is equal to the sum of the numbers to the right of the index.

If no such index exists, we should return -1. If there are multiple pivot indexes, you should return the left-most pivot index.

Example 1:

Input: 
nums = [1, 7, 3, 6, 5, 6]
Output: 3
Explanation: 
The sum of the numbers to the left of index 3 (nums[3] = 6) is equal to the sum of numbers to the right of index 3.
Also, 3 is the first index where this occurs.

Example 2:

Input: 
nums = [1, 2, 3]
Output: -1
Explanation: 
There is no index that satisfies the conditions in the problem statement.

Note:

• The length of nums will be in the range [0, 10000].
• Each element nums[i] will be an integer in the range [-1000, 1000].

A brute force solution is for each index, calculate the sum on its left and right side. This gives a O(n^2) runtime.

A better solution is to use prefix sum to avoid repeatedly computing the sum of the same subarray. 

Solution 1. O(n) runtime, O(n) space 

 1 class Solution {
 2     public int pivotIndex(int[] nums) {
 3         if(nums == null || nums.length == 0) {
 4             return -1;
 5         }
 6         int n = nums.length;
 7         int[] prefixSum = new int[n];
 8         prefixSum[0] = nums[0];
 9         for(int i = 1; i < n; i++) {
10             prefixSum[i] = prefixSum[i - 1] + nums[i];
11         }
12         int leftSum, rightSum, pivotIndex;
13         for(pivotIndex = 0; pivotIndex < n; pivotIndex++) {
14             if(pivotIndex == 0) {
15                 leftSum = 0;
16             }
17             else {
18                 leftSum = prefixSum[pivotIndex - 1];
19             }
20             rightSum = prefixSum[n - 1] - prefixSum[pivotIndex];
21             if(leftSum == rightSum) {
22                 return pivotIndex;
23             }
24         }
25         return -1;
26     }
27 }

Even better, we can solve this problem using only O(1) space without actually storing the entire prefix result. A total sum of the entire input array is all we need.

 1 class Solution {
 2     public int pivotIndex(int[] nums) {
 3         int sum = 0, left = 0;
 4         for (int i = 0; i < nums.length; i++) sum += nums[i];
 5         
 6         for (int i = 0; i < nums.length; i++) {
 7             if (i != 0) left += nums[i - 1];
 8             if (sum - left - nums[i] == left) return i;
 9         }
10         
11         return -1;
12     }
13 }