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