HackerRank Minimum Loss problem solution

In this HackerRank Minimum Loss problem solution, Lauren has a chart of distinct projected prices for a house over the next several years. She must buy the house in one year and sell it in another, and she must do so at a loss. She wants to minimize her financial loss.

Problem solution in Python.

n = int(input())
arr = list(map(int,input().split()))
di = {arr[i]:i for i in range(n)}
arr = sorted(arr)
m=10000000
for i in range(1,n):
    if di[arr[i]]<di[arr[i-1]]:
        m=min(m,arr[i]-arr[i-1])
print(m)
    

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Problem solution in Java.

import java.io.IOException;
import java.util.*;

public class Solution {

    public static void main(String[] args) throws IOException {
        new Solution().run();
    }

    private void run() throws IOException {
        Scanner in = new Scanner(System.in);

        int n = in.nextInt();
        long[] prices = new long[n];
        for (int i = 0; i < n; i++) {
            prices[i] = in.nextLong();
        }

        long minimalLoss = fasterGetMinimalLoss(n, prices);

        System.out.println(minimalLoss);
    }

    private long fasterGetMinimalLoss(int n, long[] prices) {
        TreeSet<Long> pos = new TreeSet<>();
        long minimalLoss = Long.MAX_VALUE;
        for (int i = 0; i < n; i++) {
            long endPrice = prices[i];

            Long biggerPrevious = pos.ceiling(endPrice);
            if (biggerPrevious != null) {
                long loss = biggerPrevious - endPrice;
                if (loss > 0 && loss < minimalLoss) {
                    minimalLoss = loss;
                }
            }
            pos.add(endPrice);
        }

        return minimalLoss;
    }

    private long dummyGetMinimalLoss(int n, long[] prices) {
        long minimalLoss = Long.MAX_VALUE;
        for (int i = 0; i < n; i++) {
            for (int j = i + 1; j < n; j++) {
                long loss = prices[i] - prices[j];
                if (loss > 0 && loss < minimalLoss) {
                    minimalLoss = loss;
                }
            }
        }
        return minimalLoss;
    }


}

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Problem solution in C++.

#include <iostream>
#include <vector>
#include <set>
#include <algorithm>
using namespace std;

int main()
{
   int n;
   cin >> n;
  vector<pair<long long, int> > a(n);
   for(int i = 0; i < n; i++){
    cin >> a[i].first;
    a[i].second = i;
   }
   sort(a.begin(), a.end());

    long long ans = (long long)1e18;
   for(int i = 1; i < n; i++)
   {
       if(a[i].second < a[i - 1].second)
       {
        if(ans > a[i].first - a[i - 1].first)
            ans = a[i].first - a[i - 1].first;
       }
   }
   cout << ans;
    return 0;
}

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Problem solution in C.

#include <math.h>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <assert.h>
#include <limits.h>
#include <stdbool.h>


typedef struct {
	long int val;
	int pos;
} node_t;

int cmpfunc(const void * a, const void * b)
{
	long int diff = ((node_t*)b)->val - ((node_t*)a)->val;
	if(diff < 0)
		return -1;
	else if(diff > 0)
		return 1;
	return 0;
}

int minimumLoss(int price_size, long int* price)
{
	node_t* pricenodes = malloc(sizeof(node_t)*price_size);

	for(int i=0; i<price_size; ++i) {
		pricenodes[i].val = price[i];
		pricenodes[i].pos = i;
	}

	// Reverse sort
	qsort(pricenodes, price_size, sizeof(*pricenodes), cmpfunc);

	long int curmin = -1;

	for(int i=1; i<price_size; ++i)
		if(pricenodes[i].pos > pricenodes[i-1].pos)
			if(-1 == curmin || curmin > pricenodes[i-1].val - pricenodes[i].val)
				curmin = pricenodes[i-1].val - pricenodes[i].val;

	return curmin;
}

int main() {
    int n; 
    scanf("%i", &n);
    long int *price = malloc(sizeof(long int) * n);
    for (int price_i = 0; price_i < n; price_i++) {
       scanf("%li",&price[price_i]);
    }
    int result = minimumLoss(n, price);
    printf("%d\n", result);
    return 0;
}

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