# 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]`.

## Analysis

Brute-Force

Prefix-Sum

`rightsum = Sum - nums[i] - leftsum`

• 用额外O(n)空间存prefix sum

• 只存数列总sum，和循环过程中的leftsum

``````Input
[-1,-1,-1,0,1,1]

Expected output is: 0.``````

## Solution

Prefix Sum (using array to store prefix sum) - O(n) time, O(n) space

``````class Solution {
public int pivotIndex(int[] nums) {
if (nums == null || nums.length == 0) {
return -1;
}
int[] prefixSum = new int[nums.length + 1];
prefixSum[0] = 0;
for (int i = 0; i < nums.length; i++) {
prefixSum[i + 1] = prefixSum[i] + nums[i];
}
for (int i = 1; i < prefixSum.length; i++) {
if (prefixSum[i - 1] == prefixSum[prefixSum.length - 1] - prefixSum[i]) {
return i - 1;
}
}
return -1;
}
}``````

Prefix Sum - O(n) time, O(1) space

``````class Solution {
public int pivotIndex(int[] nums) {
int sum = 0, leftsum = 0;
for (int x: nums) sum += x;
for (int i = 0; i < nums.length; ++i) {
if (leftsum == sum - leftsum - nums[i]) return i;
leftsum += nums[i];
}
return -1;
}
}``````

Brute-Force - Time: O(n^2), Space O(1)

``````class Solution {
public int pivotIndex(int[] nums) {
if (nums == null || nums.length < 3) {
return -1;
}
for (int i = 0; i < nums.length; i++) {
int left = i - 1, right = i + 1;
int sumLeft = 0, sumRight = 0;

while (left >= 0) {
sumLeft += nums[left--];
}
while (right < nums.length) {
sumRight += nums[right++];
}
if (sumLeft == sumRight) {
return i;
}
}
return -1;
}
}``````

## Reference

https://leetcode.com/articles/find-pivot-index/

https://leetcode.com/problems/find-pivot-index/discuss/109249/Java-6-liner

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