Given an array of integers nums
, write a method that returns 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.
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
Input: nums = [1,2,3]
Output: -1
Explanation:
There is no index that satisfies the conditions in the problem statement.
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
nums
will be in the range [0, 10^4]
.nums
will be in the range [-10^4, 10^4]
.nums
be empty?left_sum
to 0
.total_sum - left_sum - nums[i]
.left_sum
equals the right sum at any index i
, return i
.left_sum
by adding the value of the current element nums[i]
.-1
.def pivotIndex(nums):
total_sum = sum(nums)
left_sum = 0
for i, num in enumerate(nums):
right_sum = total_sum - left_sum - num
if left_sum == right_sum:
return i
left_sum += num
return -1
n
is the length of the array. This is because we traverse the array only once.Got blindsided by a question you didn’t expect?
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