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Minimum Loss

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  • abhiram_kankipa1
     + 0 comments
    def minimumLoss(price):
    # Write your code here
    p_dict ={}
    p_len=len(price)
    for i in range(p_len):
        p_dict[price[i]]=i
    price.sort()
    min_diff=price[1]-price[0]
    if p_len ==2:
        return min_diff
    for i in range(2,p_len):
        
        if price[i]-price[i-1]<min_diff \
            and p_dict[price[i]]< \
            p_dict[price[i-1]]:
            min_diff = price[i]-price[i-1]
    return min_diff

  • sammy27berger
     + 0 comments

    Key insight to avoid O(n^2) - after sorting the array, you can be sure that the prices you're looking for are adjacent. We're given that there is a solution - so let's call the 'optimal' buying and selling prices B and S, for **B**uy and **S**ell.

    We know B and S is a solution; therefore, pos(B) < pos(S), and B > S.

    Now for the sake of contradiction, assume B and S are not adjacent when sorted. Therefore, there must be some value X such that S < X < B. Now let's go back to the original array:

    1. If pos(X) < pos(B), then we could have formed the buy-sell pair of buying at X and selling at S. Because X is closer to S than B is, this would have been a better pair then our 'optimal' pair - a contradiction.

    2. If pos(B) < pos(X), then we could have formed the buy-sell pair of buying at B and selling at X. Because X is closer to B than S is, this would also have been better than our 'optimal' pair - another contradition.

    Therefore, we know that B and S must be adjacent when sorted. The remainder of the nlogn algorithm is left as an exercise to the reader XD

  • afzal_saiyed931
     + 1 comment
    def minimumLoss(price):
    # Write your code here
    n = len(price)
    temp = {}
    for i in range(n):
        temp[price[i]] = i
        
        
    sorted_prices = sorted(price)
    minimum_loss = None
    for i in range(n-1):
        if temp[sorted_prices[i+1]] < temp[sorted_prices[i]]:
            loss = sorted_prices[i + 1] - sorted_prices[i]
            if not minimum_loss or loss < minimum_loss:
                minimum_loss = loss
    return minimum_loss

  • gschoen
     + 0 comments

    When you need to sort the data but keep track of the original index positions, use a pair-index structure. In the code below, the 'quickSort()` routine sorts by price in non-increasing order and for elements that have have the same price, it sorts by index in increasing order.

    typedef struct {
    long long price;
    int index;
} Pair;

void quickSort(void **a, int lo, int hi);

int minimumLoss(long long* price, int n) {
    long long minLoss = 1e16;
    Pair *pair[n], p[n];
    for (int i = 0; i < n; ++i) {
        pair[i] = &p[i];
        p[i].price = price[i];
        p[i].index = i;
    }
    quickSort((void**) pair, 0, n-1);
    for (int i = 0; i < n-1; ++i) {
        if (pair[i]->index > pair[i+1]->index) continue;
        long long loss = pair[i]->price - pair[i+1]->price;
        if (loss < minLoss) minLoss = loss;
    }
    return minLoss;
}

  • mmhelloworld
     + 0 comments

    Java solution with TreeSet that provides a higher function to query the closest higher element.

        public static int minimumLoss(List<Long> prices) {
        var sortedPrices = new TreeSet<Long>();
        var minimumLoss = (long) Integer.MAX_VALUE;
        for (var price : prices) {
            var higher = sortedPrices.higher(price);
            if (higher != null) {
                minimumLoss = Math.min(minimumLoss, higher - price);
            }
            sortedPrices.add(price);
        }
        return (int) minimumLoss;
    }