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  1. Prepare
  2. Algorithms
  3. Greedy
  4. Minimum Absolute Difference in an Array
  5. Discussions

Minimum Absolute Difference in an Array

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  • vutrantrung14821
     + 0 comments
    def minimumAbsoluteDifference(arr):
    arr.sort()
    a=arr[0]-arr[1]
    b=-10000000000
    for i in range(2,n):
        b=max(b,arr[i-2]-a-arr[i])
        a=arr[i-2]-a-arr[i]
    return abs(b)

  • alban_tyrex
     + 0 comments

    Here is my c++ solution, you can find the explanation here : https://youtu.be/6SnD7A1TbJQ

    int minimumAbsoluteDifference(vector<int> arr) {
    sort(arr.begin(), arr.end());
    int result = INT_MAX;
    for(int i = 1; i < arr.size(); i++){
        int d = arr[i] - arr[i-1];
        if( d < result) result = d;
    }
    return result;
}

  • emhhacker
     + 0 comments

    My Java solution with o(n log n + n) time complexity and o(1) space:

    public static int minimumAbsoluteDifference(List<Integer> arr) {
        // goal: determine the min abs val between any two pairs in arr
        Collections.sort(arr); //sort arr ascending
        int min = Integer.MAX_VALUE;
        for(int i = 1; i < arr.size(); i++){
            int currentMin = arr.get(i) - arr.get(i - 1); 
            if(currentMin < min) min = currentMin;
        }
        return min;
    }

  • patrickgao2020
     + 0 comments

    Here is my Python solution!

    def minimumAbsoluteDifference(arr):
    arr.sort()
    differences = []
    for i in range(1, len(arr)):
        differences.append(abs(arr[i] - arr[i - 1]))
    return min(differences)

  • wilson155079
     + 0 comments
        public static int minimumAbsoluteDifference(List<Integer> arr) {
        // Sort the array
        Collections.sort(arr); 
        
        int absMin = Integer.MAX_VALUE;
        for (int i = 1; i < arr.size(); i++) {
            int absDiff = Math.abs(arr.get(i) - arr.get(i - 1));
            absMin = Math.min(absMin, absDiff);
        }
        
        return absMin;
    }