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  1. Prepare
  2. Algorithms
  3. Implementation
  4. Minimum Distances
  5. Discussions

Minimum Distances

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1393 Discussions

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  • alban_tyrex
     + 0 comments

    Here is my c++ solution, you can watch the explanation here : https://youtu.be/H8aBskFE2XI

    int minimumDistances(vector<int> a) {
    map<int, int>mp;
    int result = 1000;
    for(int i = 0; i < a.size(); i++){
        if(mp[a[i]]){
            result = min(result, i + 1 - mp[a[i]]);
        }
        mp[a[i]] = i+1;
    }
    return (result == 1000) ? -1:result;
}

  • patrickgao2020
     + 0 comments

    Here is my Python solution!

    def minimumDistances(a):
    distances = []
    if len(a) == len(set(a)):
        return -1
    for num in range(len(a)):
        if a[num] in a[num + 1:]:
            distances.append(abs(num - (a[num + 1:].index(a[num]) + num + 1)))
    return min(distances)

  • alban_tyrex
     + 0 comments

    Here is my c++ solution, you can watch the explanation here : https://youtu.be/H8aBskFE2XI

    int minimumDistances(vector<int> a) {
    map<int, int>mp;
    int result = 1000;
    for(int i = 0; i < a.size(); i++){
        if(mp[a[i]]){
            result = min(result, i + 1 - mp[a[i]]);
        }
        mp[a[i]] = i+1;
    }
    return (result == 1000) ? -1:result;
}

  • palashmondal351
     + 0 comments

    Python Here is the magic

    def minimumDistances(a):
    d = {key:abs(a.index(key) - (len(a)-1-a[::-1].index(key))) for key in set(a) if a.count(key)>1}
    val = min(d.values()) if len(d)>0 else -1
    return val

  • foxboyseraphim
     + 0 comments

    Way to do it with out nested for loops

    def minimumDistances(a):
    # Write your code here
    values = []
    for i in a:
        start = a.index(i)
        if a.count(i) >= 2:
            #d=a.index(i, start+1)
            values.append(abs(a.index(i, start+1)-start))
            print(i)
    if values:
        return min(values)
    else:
        return -1