feat(problems): add LeetCode array problem solutions

This commit is contained in:
Prad Nukala
2026-08-18 14:34:02 -04:00
parent 9e8117fcf6
commit abe86648f8
4 changed files with 252 additions and 0 deletions
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/*
* 1413. Minimum Value to Get Positive Step by Step Sum
* Difficulty: Easy
* https://leetcode.com/problems/minimum-value-to-get-positive-step-by-step-sum/
*
* ──────────────────────────────────────────────────
*
* Given an array of integers nums, you start with an initial positive
* value startValue.
*
* In each iteration, you calculate the step by step sum of startValue
* plus elements in nums (from left to right).
*
* Return the minimum positive value of startValue such that the step by
* step sum is never less than 1.
*
*
*
* Example 1:
*
* Input: nums = [-3,2,-3,4,2]
* Output: 5
* Explanation: If you choose startValue = 4, in the third iteration
* your step by step sum is less than 1.
* step by step sum
* startValue = 4 | startValue = 5 | nums
* (4 -3 ) = 1 | (5 -3 ) = 2 | -3
* (1 +2 ) = 3 | (2 +2 ) = 4 | 2
* (3 -3 ) = 0 | (4 -3 ) = 1 | -3
* (0 +4 ) = 4 | (1 +4 ) = 5 | 4
* (4 +2 ) = 6 | (5 +2 ) = 7 | 2
*
* Example 2:
*
* Input: nums = [1,2]
* Output: 1
* Explanation: Minimum start value should be positive.
*
* Example 3:
*
* Input: nums = [1,-2,-3]
* Output: 5
*
*
*
* Constraints:
*
* • 1 <= nums.length <= 100
*
* • -100 <= nums[i] <= 100
*/
/**
* @param {number[]} nums
* @return {number}
*/
var minStartValue = function(nums) {
let prefix = [nums[0]];
// Make prefix sum start at 1 after initializing seed value
for (let i = 1; i < nums.length; i++){
prefix.push(prefix[i - 1] + nums[i]);
}
let min = Math.min(...prefix);
return Math.max(1, 1 - min);
};
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/*
* 1480. Running Sum of 1d Array
* Difficulty: Easy
* https://leetcode.com/problems/running-sum-of-1d-array/
*
* ──────────────────────────────────────────────────
*
* Given an array nums. We define a running sum of an array as
* runningSum[i] = sum(nums[0]&hellip;nums[i]).
*
* Return the running sum of nums.
*
*
*
* Example 1:
*
* Input: nums = [1,2,3,4]
* Output: [1,3,6,10]
* Explanation: Running sum is obtained as follows: [1, 1+2, 1+2+3,
* 1+2+3+4].
*
* Example 2:
*
* Input: nums = [1,1,1,1,1]
* Output: [1,2,3,4,5]
* Explanation: Running sum is obtained as follows: [1, 1+1, 1+1+1,
* 1+1+1+1, 1+1+1+1+1].
*
* Example 3:
*
* Input: nums = [3,1,2,10,1]
* Output: [3,4,6,16,17]
*
*
*
* Constraints:
*
* • 1 <= nums.length <= 1000
*
* • -10^6 <= nums[i] <= 10^6
*/
/**
* @param {number[]} nums
* @return {number[]}
*/
var runningSum = function(nums) {
let prefix = [nums[0]];
for (let i = 1; i < nums.length; i++){
prefix.push(prefix[i - 1] + nums[i]);
}
return prefix;
};
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/*
* 2090. K Radius Subarray Averages
* Difficulty: Medium
* https://leetcode.com/problems/k-radius-subarray-averages/
*
* ──────────────────────────────────────────────────
*
* You are given a 0-indexed array nums of n integers, and an integer k.
*
* The k-radius average for a subarray of nums centered at some index i
* with the radius k is the average of all elements in nums between the
* indices i - k and i + k (inclusive). If there are less than k elements
* before or after the index i, then the k-radius average is -1.
*
* Build and return an array avgs of length n where avgs[i] is the
* k-radius average for the subarray centered at index i.
*
* The average of x elements is the sum of the x elements divided by x,
* using integer division. The integer division truncates toward zero,
* which means losing its fractional part.
*
* • For example, the average of four elements 2, 3, 1, and 5 is (2 + 3
* + 1 + 5) / 4 = 11 / 4 = 2.75, which truncates to 2.
*
*
*
* Example 1:
*
* Input: nums = [7,4,3,9,1,8,5,2,6], k = 3
* Output: [-1,-1,-1,5,4,4,-1,-1,-1]
* Explanation:
* - avg[0], avg[1], and avg[2] are -1 because there are less than k
* elements before each index.
* - The sum of the subarray centered at index 3 with radius 3 is: 7 + 4
* + 3 + 9 + 1 + 8 + 5 = 37.
* Using integer division, avg[3] = 37 / 7 = 5.
* - For the subarray centered at index 4, avg[4] = (4 + 3 + 9 + 1 + 8 +
* 5 + 2) / 7 = 4.
* - For the subarray centered at index 5, avg[5] = (3 + 9 + 1 + 8 + 5 +
* 2 + 6) / 7 = 4.
* - avg[6], avg[7], and avg[8] are -1 because there are less than k
* elements after each index.
*
* Example 2:
*
* Input: nums = [100000], k = 0
* Output: [100000]
* Explanation:
* - The sum of the subarray centered at index 0 with radius 0 is:
* 100000.
* avg[0] = 100000 / 1 = 100000.
*
* Example 3:
*
* Input: nums = [8], k = 100000
* Output: [-1]
* Explanation:
* - avg[0] is -1 because there are less than k elements before and
* after index 0.
*
*
*
* Constraints:
*
* • n == nums.length
*
* • 1 <= n <= 10^5
*
* • 0 <= nums[i], k <= 10^5
*/
/**
* @param {number[]} nums
* @param {number} k
* @return {number[]}
*/
var getAverages = function(nums, k) {
};
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/*
* 238. Product of Array Except Self
* Difficulty: Medium
* https://leetcode.com/problems/product-of-array-except-self/
*
* ──────────────────────────────────────────────────
*
* Given an integer array nums, return an array answer such that
* answer[i] is equal to the product of all the elements of nums except
* nums[i].
*
* The product of any prefix or suffix of nums is guaranteed to fit in a
* 32-bit integer.
*
* You must write an algorithm that runs in O(n) time and without using
* the division operation.
*
*
*
* Example 1:
*
* Input: nums = [1,2,3,4]
* Output: [24,12,8,6]
*
* Example 2:
*
* Input: nums = [-1,1,0,-3,3]
* Output: [0,0,9,0,0]
*
*
*
* Constraints:
*
* • 2 <= nums.length <= 10^5
*
* • -30 <= nums[i] <= 30
*
* • The input is generated such that answer[i] is guaranteed to fit in
* a 32-bit integer.
*
*
*
* Follow up: Can you solve the problem in O(1) extra space complexity?
* (The output array does not count as extra space for space complexity
* analysis.)
*/
/**
* @param {number[]} _nums
* @return {number[]}
*/
var productExceptSelf = function(_nums
};