Final Prices With a Special Discount in a Shop - Array - Easy - LeetCode
💻 coding

Final Prices With a Special Discount in a Shop - Array - Easy - LeetCode

2 min read 457 words
2 min read
ShareWhatsAppPost on X
  • 1The algorithm calculates final prices by applying discounts based on future prices in the array.
  • 2If a future price is less than or equal to the current price, the discount is applied.
  • 3The solution has a time complexity of O(n^2) and a space complexity of O(n).

AI-generated summary · May not capture all nuances

Key Insight
AskGif

"The algorithm calculates final prices by applying discounts based on future prices in the array."

Final Prices With a Special Discount in a Shop - Array - Easy - LeetCode

Given the array prices where prices[i] is the price of the ith item in a shop. There is a special discount for items in the shop, if you buy the ith item, then you will receive a discount equivalent to prices[j] where j is the minimum index such that j > i and prices[j] <= prices[i], otherwise, you will not receive any discount at all.

Return an array where the ith element is the final price you will pay for the ith item of the shop considering the special discount.

Example 1:

Input: prices = [8,4,6,2,3]

Output: [4,2,4,2,3]

Explanation:

For item 0 with price[0]=8 you will receive a discount equivalent to prices[1]=4, therefore, the final price you will pay is 8 - 4 = 4.

For item 1 with price[1]=4 you will receive a discount equivalent to prices[3]=2, therefore, the final price you will pay is 4 - 2 = 2.

For item 2 with price[2]=6 you will receive a discount equivalent to prices[3]=2, therefore, the final price you will pay is 6 - 2 = 4.

For items 3 and 4 you will not receive any discount at all.

Example 2:

Input: prices = [1,2,3,4,5]

Output: [1,2,3,4,5]

Explanation: In this case, for all items, you will not receive any discount at all.

Example 3:

Input: prices = [10,1,1,6]

Output: [9,0,1,6]

Constraints:

1 <= prices.length <= 500

1 <= prices[i] <= 10^3

Solution:

using System;
using System.Collections.Generic;
using System.Text;

namespace LeetCode.AskGif.Easy.Array
{
 public class FinalPricesSoln
 {
 public int[] FinalPrices(int[] prices)
 {
 var discount = new int[prices.Length];
 for (int i = 0; i < prices.Length; i++)
 {
 discount[i] = prices[i];
 for (int j = i+1; j < prices.Length; j++)
 {
 if(prices[j] <= prices[i])
 {
 discount[i] = prices[i] - prices[j];
 break;
 }
 }
 }

 return discount;
 }
 }
}

Time Complexity: O(n^2)

Space Complexity: O(n) To store results.

Unit Tests:

using LeetCode.AskGif.Easy.Array;
using Microsoft.VisualStudio.TestTools.UnitTesting;
using System;
using System.Collections.Generic;
using System.Text;

namespace CodingUnitTest.Easy.Array
{
 [TestClass]
 public class FinalPricesSolnTests
 {
 [TestMethod]
 public void FinalPricesSoln_First()
 {
 var prices = new[] { 8, 4, 6, 2, 3 };
 var expected = new int[] { 4, 2, 4, 2, 3 };

 var res = new FinalPricesSoln().FinalPrices(prices);
 AreEqual(expected, res);
 }

 [TestMethod]
 public void FinalPricesSoln_Second()
 {
 var prices = new[] { 1, 2, 3, 4, 5 };
 var expected = new int[] { 1, 2, 3, 4, 5 };

 var res = new FinalPricesSoln().FinalPrices(prices);
 AreEqual(expected, res);
 }

 [TestMethod]
 public void FinalPricesSoln_Third()
 {
 var prices = new[] { 10, 1, 1, 6 };
 var expected = new int[] { 9, 0, 1, 6 };

 var res = new FinalPricesSoln().FinalPrices(prices);
 AreEqual(expected, res);
 }

 private void AreEqual(int[] res, int[] output)
 {
 Assert.AreEqual(res.Length, output.Length);
 for (int i = 0; i < res.Length; i++)
 {
 Assert.AreEqual(res[i], output[i]);
 }
 }
 }
}

Enjoyed this article?

Share it with someone who'd find it useful.

ShareWhatsAppPost on X

AskGif

Published on 17 June 2020 · 2 min read · 457 words

Part of AskGif Blog · coding

You might also like