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  1. Minimum Loss 1
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Minimum Loss 1

  • robjrm
     + 0 comments

    Since similar solutions rely on either a sorted array or using binary search n times where n is the length of the array, the time complexity comes out to O(nlogn) either way, I figured this is more intuitive:

    def minimumLoss(price):
    # Write your code here
    p_to_i = {}
    for i,p in enumerate(price):
        p_to_i[p] = i
    
    price.sort()
    min_value = math.inf
    for i in range(1,len(price)):
        if p_to_i[price[i-1]] > p_to_i[price[i]]:
            min_value = min(min_value,price[i] - price[i-1])
    return min_value