# Random Wits

Life is too short for a diary

\$ random wits
├── tags

├── bookshelf

├── resources

├── quotes

├── habits

Tue 02 Nov 2021

## Coin Change solution leetcode

It's one of the most popular questions on leetcode that seems very easy at first. Coin change is a classic dynamic programming problem. I will proceed with an obvious (albeit wrong) solution and subsequently proceed to an efficient correct solution.

## Problem Statement

You are given an integer array coins representing coins of different denominations and an integer amount representing a total amount of money. Return the fewest number of coins that you need to make up that amount. If that amount of money cannot be made up by any combination of the coins, return -1. You may assume that you have an infinite number of each kind of coin.

## Greedy Solution

My first instinct is to sort the coins array. Pick the largest coin first and then subtract the largest possible value from the amount. Subsequently, proceed with smaller denominations while keeping track of the count.

However this solution fails the following test case as the minimum sequence is [8, 4]

coins = [1, 2, 4, 8, 9]
amount = 12

Output : 3
Expected: 2


## Dynamic Programming

We have to find some subproblem that will help in solving coin-change problem. Given an amount n, we want to generate an exact change using the fewest number of coins of denominations d1 < d2 < … < dm.

Let Dp(n) represent the minimum number of coins required for a given amount n. Coins dj can be added to amount n - dj only if dj <= n and 0 <= j <= n -1 wher Dp(0) is 0.

\begin{equation} Dp(n) = \min_{j=0}^{d_j<=n} Dp(n - D_j) + 1 \end{equation}

Let us proceed with following test case

coins = [1, 2, 5]
amount = 11


We can vary the amount as i from 0 to amount. The value of Dp[i] for each iteration becomes

0 1 2 3 4 5 6 7 8 9 10 11
i = 0 0 12 12 12 12 12 12 12 12 12 12 12
i = 1 0 1 12 12 12 12 12 12 12 12 12 12
i = 2 0 1 1 12 12 12 12 12 12 12 12 12
i = 3 0 1 1 2 12 12 12 12 12 12 12 12
i = 4 0 1 1 2 2 12 12 12 12 12 12 12
i = 5 0 1 1 2 2 1 12 12 12 12 12 12
i = 6 0 1 1 2 2 1 2 12 12 12 12 12
i = 7 0 1 1 2 2 1 2 2 12 12 12 12
i = 8 0 1 1 2 2 1 2 2 3 12 12 12
i = 9 0 1 1 2 2 1 2 2 3 3 12 12
i = 10 0 1 1 2 2 1 2 2 3 3 2 12
i = 11 0 1 1 2 2 1 2 2 3 3 2 3

## Complexity

Time Complexity is $$\theta(n * m)$$, Space complexity is $$\theta(n)$$