Problem
You are given an integer array nums
and an integer x
. In one operation, you can either remove the leftmost or the rightmost element from the array nums
and subtract its value from x
. Note that this modifies the array for future operations.
Return **the *minimum number* of operations to reduce **x
**to *exactly 0
*if it is possible, otherwise, return **-1
.
Example 1:
Input: nums = [1,1,4,2,3], x = 5
Output: 2
Explanation: The optimal solution is to remove the last two elements to reduce x to zero.
Example 2:
Input: nums = [5,6,7,8,9], x = 4
Output: -1
Example 3:
Input: nums = [3,2,20,1,1,3], x = 10
Output: 5
Explanation: The optimal solution is to remove the last three elements and the first two elements (5 operations in total) to reduce x to zero.
Constraints:
1 <= nums.length <= 105
1 <= nums[i] <= 104
1 <= x <= 109
Solution
/**
* @param {number[]} nums
* @param {number} x
* @return {number}
*/
var minOperations = function(nums, x) {
var leftSumMap = { 0: 0 };
var rightSumMap = { 0: 0 };
var leftSum = 0;
var rightSum = 0;
var min = Number.MAX_SAFE_INTEGER;
for (var i = 0; i < nums.length; i++) {
leftSum += nums[i];
rightSum += nums[nums.length - 1 - i];
leftSumMap[leftSum] = i + 1;
rightSumMap[rightSum] = i + 1;
if (rightSumMap[x - leftSum] !== undefined && (i + 1 + rightSumMap[x - leftSum]) <= nums.length) {
min = Math.min(min, i + 1 + rightSumMap[x - leftSum]);
}
if (leftSumMap[x - rightSum] !== undefined && (i + 1 + leftSumMap[x - rightSum]) <= nums.length) {
min = Math.min(min, i + 1 + leftSumMap[x - rightSum]);
}
}
return min === Number.MAX_SAFE_INTEGER ? -1 : min;
};
Explain:
nope.
Complexity:
- Time complexity : O(n).
- Space complexity : O(n * n).