HackerRank Minimum Loss Problem Solution

In this post, we will solve 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.
Example
price = [20, 15, 8, 2, 12]
Her minimum loss is incurred by purchasing in year 2 at price[1] = 15 and reselling in year 5 at price[4] = 12. Return 15- 12 = 3.

Function Description

Complete the minimumLoss function in the editor below.

minimumLoss has the following parameter(s):

int price[n]: home prices at each year

Returns

int: the minimum loss possible

Input Format

The first line contains an integer n, the number of years of house data.
The second line contains n space-separated long integers that describe each price[i].

Sample Input 0

3
5 10 3

Sample Output 0

Explanation 0
Lauren buys the house in year 1 at price [0] = 5 and sells it in year 3 at price[2] = 3 for a minimal loss of 5-32.

Sample Input 1

5
20 7 8 2 5

Sample Output 1

Explanation 1
Lauren buys the house in year 2 at price[1] = 7 and sells it in year 5 at price[4] = 5 for a minimal loss of 7-5 = 2.

HackerRank Minimum Loss Problem Solution

Minimum Loss C Solution

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

struct node {
	long long val;
	struct node *left;
	struct node *right;
};

typedef struct node node_t;

static node_t *
init_tree(long long val)
{
	node_t *tree = NULL;
	tree = calloc(1, sizeof(node_t));
	tree->val = val;
	return tree;
}

static void
insert(node_t *tree, node_t *t, long long *min)
{
	if (tree->val > t->val) {
		*min = tree->val;
		if (tree->left) {
			return insert(tree->left, t, min);
		} else {
			tree->left = t;
			return;
		}
	} else {
		if (tree->right) {
			return insert(tree->right, t, min);
		} else {
			tree->right = t;
		}
	}
	return;
}

int main() {

    long long unsigned min_loss = ULONG_MAX;
    long long N;
    long long i, val, min;
    node_t *temp = NULL;
    node_t *tree = NULL;

    scanf("%lld", &N);
    for (i = 0; i < N; i++) {
	scanf("%lld", &val);
	min = -1;
	if (i == 0) {
		tree = init_tree(val);
		continue;
	}
	temp = calloc(1, sizeof(node_t));
	temp->val = val;
	insert(tree, temp, &min);
	if ((min != -1) && (min_loss > (min - val))) {
		min_loss = min - val;
	}
    }
    printf("%lld\n", min_loss);
    return 0;
}

Minimum Loss C++ Solution

#include <bits/stdc++.h>

using namespace std;

#define on(s, j) ((s >> j) & 1)
#define lowbit(s) (s & (-s))
#define dbg(x) cerr << #x << " : " << (x) << " "
#define print(x) cerr << (x) << " "
#define newline cerr << "\n"

typedef long long ll;

inline int lsb(ll x) {return    __builtin_ctzll(x);}
inline int gi() {int x; scanf("%d", &x);  return x;}
inline ll gll() {ll x; scanf("%lld", &x); return x;}
inline int msb(ll x) {return 63-__builtin_clzll(x);}

template <class T>
void debug(T a, T b) {
  cerr << "[";
  for (T i = a; i != b; ++i) {
    if (i != a) cerr << ", ";
    cerr << *i;
  }
  cerr << "]\n";
}

const int N = 200010;

ll p[N];

int main() {
  // freopen("input.in", "r", stdin);
  // freopen("output.out", "w", stdout);

  int n = gi();
  for (int i = 0; i < n; i++) {
    p[i] = gll();
  }
  ll ans = (ll)1e17;
  set <ll> st;
  for (int i = n - 1; i >= 0; i--) {
    if (i == n - 1) {
      st.insert(p[i]);
    } else {
      set <ll> :: iterator it = st.lower_bound(p[i]);
      if (it != st.begin()) {
        it--;
        ans = min(ans, p[i] - *it);
      }
      st.insert(p[i]);
    }
  }
  printf("%lld\n", ans);

}

Minimum Loss C Sharp Solution

using System;

using System.Collections.Generic;

using System.IO;

using System.Linq;

class Solution
{
    static void Main(String[] args)
    {

        int n = Int32.Parse(Console.ReadLine());

        double[] price = Console.ReadLine().Split(' ').Select(double.Parse).ToArray();

        Console.WriteLine(DisplayMinLoss(price));

    }



    private static double DisplayMinLoss(double[] price)
    {
        List<double> sortedList = price.ToList();
        sortedList.Sort();
        double minimum = sortedList.Sum();
        for (int i = 0; i < sortedList.Count - 1; i++)
        {
            if ((sortedList[i + 1] - sortedList[i]) < minimum && Array.IndexOf(price, sortedList[i + 1]) < Array.IndexOf(price, sortedList[i]))
            {

                minimum = sortedList[i + 1] - sortedList[i];

            }            
        }

        return minimum;

    }

}

Minimum Loss Java Solution

import java.io.*;
import java.math.*;
import java.security.*;
import java.text.*;
import java.util.*;
import java.util.concurrent.*;
import java.util.function.*;
import java.util.regex.*;
import java.util.stream.*;
import static java.util.stream.Collectors.joining;
import static java.util.stream.Collectors.toList;
class Result {
public static int minimumLoss(List<Long> price) {
    // Write your code here
 
 HashMap<Long,Integer> h=new HashMap<>();
 for(int i=0;i<price.size();i++)
{
    
    h.put(price.get(i),i);
}
Collections.sort(price);
long min=Integer.MAX_VALUE;
for(int i=1;i<price.size();i++)
{
   
    if(h.get(price.get(i))<h.get(price.get(i-1)) && (price.get(i)-price.get(i-1))<min)
    {
        min=price.get(i)-price.get(i-1);
    }
    
}
return (int)min;
    }

}

public class Solution {
    public static void main(String[] args) throws IOException {
        BufferedReader bufferedReader = new BufferedReader(new InputStreamReader(System.in));
        BufferedWriter bufferedWriter = new BufferedWriter(new FileWriter(System.getenv("OUTPUT_PATH")));

        int n = Integer.parseInt(bufferedReader.readLine().trim());

        List<Long> price = Stream.of(bufferedReader.readLine().replaceAll("\\s+$", "").split(" "))
            .map(Long::parseLong)
            .collect(toList());

        int result = Result.minimumLoss(price);

        bufferedWriter.write(String.valueOf(result));
        bufferedWriter.newLine();

        bufferedReader.close();
        bufferedWriter.close();
    }
}

Minimum Loss JavaScript Solution



function processData(input) {
    var x = input.split("\n");
    var n = Number(x[0]);
    var s = x[1].split(' ');
    var a = {};
    var loss = Infinity;
    var matched = false;
    for (var i=0; i < s.length; i++) {
        s[i] = Number(s[i]);
        if (a[s[i]] !== undefined) {
            matched = true;
            break;
        }
        a[s[i]] = i;
    }
    if (matched) {
        return 0;
    }
    var compr = function (x, y) {
        return y-x;
    }
    s.sort(compr);
    var x,y;
    for (var i=1; i < s.length; i++) {
        x = s[i-1];
        y = s[i];
        if (a[x] < a[y]) {
            loss = (loss > x - y) ? x-y : loss;
        }
    }
    console.log(loss);
} 

process.stdin.resume();
process.stdin.setEncoding("ascii");
_input = "";
process.stdin.on("data", function (input) {
    _input += input;
});

process.stdin.on("end", function () {
   processData(_input);
});

Minimum Loss Python Solution

n =  int(input().strip())
numbers = list(map(int,input().strip().split()))

nums = list(numbers)
nums.sort()
minCost = sum(nums)
for i in range(1,n):
    if (nums[i]-nums[i-1] < minCost)  and (numbers.index(nums[i]) < numbers.index(nums[i-1])):
        minCost = nums[i]-nums[i-1]
print(minCost)

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