# 724. Find Pivot Index

# Problem

Given an array of integers `nums`

, calculate the **pivot index** of this array.

The **pivot index** is the index where the sum of all the numbers **strictly** to the left of the index is equal to the sum of all the numbers **strictly** to the index's right.

If the index is on the left edge of the array, then the left sum is `0`

because there are no elements to the left. This also applies to the right edge of the array.

Return *the leftmost pivot index*. If no such index exists, return

`-1`

.**Example 1:**

**Input:** nums = [1,7,3,6,5,6] **Output:** 3 **Explanation:**The pivot index is 3. Left sum = nums[0] + nums[1] + nums[2] = 1 + 7 + 3 = 11 Right sum = nums[4] + nums[5] = 5 + 6 = 11

**Example 2:**

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

**Example 3:**

**Input:** nums = [2,1,-1] **Output:** 0 **Explanation:**The pivot index is 0. Left sum = 0 (no elements to the left of index 0) Right sum = nums[1] + nums[2] = 1 + -1 = 0

**Constraints:**

`1 <= nums.length <= 10`

^{4}`-1000 <= nums[i] <= 1000`

**Note:** This question is the same as 1991: https://leetcode.com/problems/find-the-middle-index-in-array/

# Solution

```
class Solution(object):
def pivotIndex(self, nums):
"""
:type nums: List[int]
:rtype: int
"""
for i in range(0, len(nums)):
if sum(nums[0:i]) == sum(nums[i + 1:]):
return i
return -1
```