47 lines
1.4 KiB
Go
47 lines
1.4 KiB
Go
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/*
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Problem Statement : Given an array of integers nums and an integer target, return indices of
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the two numbers such that they add up to target.
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You may assume that each input would have exactly one solution, and you may
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not use the same element twice.
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Input: An array of integers and a target (int)
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Output: array of indexes of len(2) with sum of element at that index equal to target or nil
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*/
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package arrays
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/*
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Using Brute Force : For every element check for another element if it exist in the array such that sum of
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both the element is equals to the target
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Time Complexity : O(n^2)
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*/
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func twoSum(arr []int, target int) []int {
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for i := 0; i < len(arr); i++ {
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for j := i + 1; j < len(arr); j++ {
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if arr[i]+arr[j] == target {
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return []int{i, j}
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}
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}
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}
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return nil
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}
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/*
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Using Map := While traversing every element add the element as key and its position as its value in a map
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Check the required value (i.e target - arr[i]) in the map
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If the map contains the required value then we have two elements with the required sum and
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return the positions.
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Time Complexity : O(n)
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*/
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func twoSumEfficient(arr []int, target int) []int {
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m := make(map[int]int)
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for i := 0; i < len(arr); i++ {
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compliment := target - arr[i]
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if _, ok := m[compliment]; ok {
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return []int{m[compliment], i}
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}
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m[arr[i]] = i
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}
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return []int{}
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}
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