Minimum Distances HackerRank Solution ( Better Logic )

Minimum Distances 2.py

def minimumDistances(a):
    n=len(a)
    minimum=n+1      #Initializing with n+1 because if no updates take place we should return -1 as there are no matching pairs
    # Note:- If any matching pairs is found the distance would be less than n
    indexes=dict()      #Creating an empty dictionary
    for i in range(n):
        if a[i] in indexes:    #If elem is present, append its index in its list
            indexes[a[i]].append(i)
        else: 		 #If elem is not present,initialize it by a list with t's index 
            indexes[a[i]]=[i]
    for i in indexes:           #For every element in the dictionary
        length=len(indexes[i])
        if n>1:       #We only have to traverse if there are more than 1 element
            for j in range(1,length): #Traverse & check the distance between two indexes
                if indexes[i][j]-indexes[i][j-1] < minimum: #Update if required
                    minimum=indexes[i][j]-indexes[i][j-1]
    if minimum==n+1:        #If no matching elements return -1
        return -1
    return minimum