Find The Middle Index In Array
Problem​
Given a 0-indexed
integer array nums, find the leftmost middleIndex (i.e., the smallest amongst all the possible ones)
.
A middleIndex is an index where nums[0] + nums[1] + ... + nums[middleIndex-1] == nums[middleIndex+1] + nums[middleIndex+2] + ... + nums[nums.length-1].
If middleIndex == 0
, the left side sum is considered to be 0
. Similarly, if middleIndex == nums.length - 1
, the right side sum is considered to be 0
.
Return the leftmost middleIndex that satisfies the condition, or -1 if there is no such index.
Examples​
Example 1:
Input: nums = [2,3,-1,8,4]
Output: 3
Explanation: The sum of the numbers before index 3 is: 2 + 3 + -1 = 4 The sum of the numbers after index 3 is: 4 = 4
Example 2:
Input: nums = [1,-1,4]
Output: 2
Explanation: The sum of the numbers before index 2 is: 1 + -1 = 0 The sum of the numbers after index 2 is: 0
Example 3:
Input: nums = [2,5]
Output: -1
Explanation: There is no valid middleIndex.
Constraints​
1 <= nums.length <= 100
-1000 <= nums[i] <= 1000
Approach​
-
Initialize prefix to 0 and suffix to the sum of all elements in nums.
-
Iterate through the elements of nums:
- For each element, subtract it from suffix.
- Check if prefix is equal to suffix. If true, return the current index i.
- Add the current element to prefix.
- If no such index is found, return -1.
Solution for Remove Linked List Elements​
Java Solution​
class Solution {
public int findMiddleIndex(int[] nums) {
int prefix = 0;
int suffix = Arrays.stream(nums).sum();
for (int i = 0; i < nums.length; ++i) {
suffix -= nums[i];
if (prefix == suffix)
return i;
prefix += nums[i];
}
return -1;
}
}
C++ solution​
class Solution {
public:
int findMiddleIndex(vector<int>& nums) {
int prefix = 0;
int suffix = accumulate(nums.begin(), nums.end(), 0);
for (int i = 0; i < nums.size(); ++i) {
suffix -= nums[i];
if (prefix == suffix)
return i;
prefix += nums[i];
}
return -1;
}
};
Python Solution​
class Solution:
def findMiddleIndex(self, nums: List[int]) -> int:
prefix = 0
suffix = sum(nums)
for i, num in enumerate(nums):
suffix -= num
if prefix == suffix:
return i
prefix += num
return -1
Complexity Analysis​
Time Complexity: O(n)
Space Complexity: O(1)
References​
LeetCode Problem: find-the-middle-index-in-array