GoLang: Check If N and Its Double Exist (Hash Map)

Given an array arr of integers, check if there exists two integers N and M such that N is the double of M ( i.e. N = 2 * M).

More formally check if there exists two indices i and j such that :
i != j
0 <= i, j < arr.length
arr[i] == 2 * arr[j]

Example 1:
Input: arr = [10,2,5,3]
Output: true
Explanation: N = 10 is the double of M = 5,that is, 10 = 2 * 5.

Example 2:
Input: arr = [7,1,14,11]
Output: true
Explanation: N = 14 is the double of M = 7,that is, 14 = 2 * 7.

Example 3:
Input: arr = [3,1,7,11]
Output: false
Explanation: In this case does not exist N and M, such that N = 2 * M.

Constraints:
2 <= arr.length <= 500
-10^3 <= arr[i] <= 10^3

Hints:
Loop from i = 0 to arr.length, maintaining in a hashTable the array elements from [0, i – 1].
On each step of the loop check if we have seen the element 2 * arr[i] so far or arr[i] / 2 was seen if arr[i] % 2 == 0.

GoLang: Check If N and Its Double Exist via Hash Maps

We can count the number of frequencies for each element in the array and store them in a hash map. In Golang, we can iterate an array using for key, val := range array.

Then, we can iterate the map using for key, value := range map syntax (which is basically the same as iterating over the array indices/values).

The zero has to be treated separately as its double is still zero.

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func checkIfExist(arr []int) bool {
    var data = make(map[int]int)
    for _, n := range arr {
        data[n] ++
    }
    for key, value := range data {
        if key == 0 {
            if value > 1 {
                return true
            }
        } else if _, ok := data[key * 2]; ok {
            return true
        }
    }
    return false
}

func checkIfExist(arr []int) bool {
    var data = make(map[int]int)
    for _, n := range arr {
        data[n] ++
    }
    for key, value := range data {
        if key == 0 {
            if value > 1 {
                return true
            }
        } else if _, ok := data[key * 2]; ok {
            return true
        }
    }
    return false
}

Time complexity is O(N) and space complexity is also O(N) as we are using a map to store the numbers and occurences.

–EOF (The Ultimate Computing & Technology Blog) —