Description#
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.
Example 1:
Input: coins = [1,2,5], amount = 11
Output: 3
Explanation: 11 = 5 + 5 + 1
Example 2:
Input: coins = [2], amount = 3
Output: -1
Example 3:
Input: coins = [1], amount = 0
Output: 0
Constraints:
1 <= coins.length <= 121 <= coins[i] <= 231 - 10 <= amount <= 104
Solutions#
Solution 1#
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| class Solution:
def coinChange(self, coins: List[int], amount: int) -> int:
m, n = len(coins), amount
f = [[inf] * (n + 1) for _ in range(m + 1)]
f[0][0] = 0
for i, x in enumerate(coins, 1):
for j in range(n + 1):
f[i][j] = f[i - 1][j]
if j >= x:
f[i][j] = min(f[i][j], f[i][j - x] + 1)
return -1 if f[m][n] >= inf else f[m][n]
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| class Solution {
public int coinChange(int[] coins, int amount) {
final int inf = 1 << 30;
int m = coins.length;
int n = amount;
int[][] f = new int[m + 1][n + 1];
for (var g : f) {
Arrays.fill(g, inf);
}
f[0][0] = 0;
for (int i = 1; i <= m; ++i) {
for (int j = 0; j <= n; ++j) {
f[i][j] = f[i - 1][j];
if (j >= coins[i - 1]) {
f[i][j] = Math.min(f[i][j], f[i][j - coins[i - 1]] + 1);
}
}
}
return f[m][n] >= inf ? -1 : f[m][n];
}
}
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| class Solution {
public:
int coinChange(vector<int>& coins, int amount) {
int m = coins.size(), n = amount;
int f[m + 1][n + 1];
memset(f, 0x3f, sizeof(f));
f[0][0] = 0;
for (int i = 1; i <= m; ++i) {
for (int j = 0; j <= n; ++j) {
f[i][j] = f[i - 1][j];
if (j >= coins[i - 1]) {
f[i][j] = min(f[i][j], f[i][j - coins[i - 1]] + 1);
}
}
}
return f[m][n] > n ? -1 : f[m][n];
}
};
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| func coinChange(coins []int, amount int) int {
m, n := len(coins), amount
f := make([][]int, m+1)
const inf = 1 << 30
for i := range f {
f[i] = make([]int, n+1)
for j := range f[i] {
f[i][j] = inf
}
}
f[0][0] = 0
for i := 1; i <= m; i++ {
for j := 0; j <= n; j++ {
f[i][j] = f[i-1][j]
if j >= coins[i-1] {
f[i][j] = min(f[i][j], f[i][j-coins[i-1]]+1)
}
}
}
if f[m][n] > n {
return -1
}
return f[m][n]
}
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| function coinChange(coins: number[], amount: number): number {
const m = coins.length;
const n = amount;
const f: number[][] = Array(m + 1)
.fill(0)
.map(() => Array(n + 1).fill(1 << 30));
f[0][0] = 0;
for (let i = 1; i <= m; ++i) {
for (let j = 0; j <= n; ++j) {
f[i][j] = f[i - 1][j];
if (j >= coins[i - 1]) {
f[i][j] = Math.min(f[i][j], f[i][j - coins[i - 1]] + 1);
}
}
}
return f[m][n] > n ? -1 : f[m][n];
}
|
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| impl Solution {
pub fn coin_change(coins: Vec<i32>, amount: i32) -> i32 {
let n = amount as usize;
let mut f = vec![n + 1; n + 1];
f[0] = 0;
for &x in &coins {
for j in x as usize..=n {
f[j] = f[j].min(f[j - (x as usize)] + 1);
}
}
if f[n] > n {
-1
} else {
f[n] as i32
}
}
}
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| /**
* @param {number[]} coins
* @param {number} amount
* @return {number}
*/
var coinChange = function (coins, amount) {
const m = coins.length;
const n = amount;
const f = Array(m + 1)
.fill(0)
.map(() => Array(n + 1).fill(1 << 30));
f[0][0] = 0;
for (let i = 1; i <= m; ++i) {
for (let j = 0; j <= n; ++j) {
f[i][j] = f[i - 1][j];
if (j >= coins[i - 1]) {
f[i][j] = Math.min(f[i][j], f[i][j - coins[i - 1]] + 1);
}
}
}
return f[m][n] > n ? -1 : f[m][n];
};
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Solution 2#
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| class Solution:
def coinChange(self, coins: List[int], amount: int) -> int:
n = amount
f = [0] + [inf] * n
for x in coins:
for j in range(x, n + 1):
f[j] = min(f[j], f[j - x] + 1)
return -1 if f[n] >= inf else f[n]
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| class Solution {
public int coinChange(int[] coins, int amount) {
final int inf = 1 << 30;
int n = amount;
int[] f = new int[n + 1];
Arrays.fill(f, inf);
f[0] = 0;
for (int x : coins) {
for (int j = x; j <= n; ++j) {
f[j] = Math.min(f[j], f[j - x] + 1);
}
}
return f[n] >= inf ? -1 : f[n];
}
}
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| class Solution {
public:
int coinChange(vector<int>& coins, int amount) {
int n = amount;
int f[n + 1];
memset(f, 0x3f, sizeof(f));
f[0] = 0;
for (int x : coins) {
for (int j = x; j <= n; ++j) {
f[j] = min(f[j], f[j - x] + 1);
}
}
return f[n] > n ? -1 : f[n];
}
};
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| func coinChange(coins []int, amount int) int {
n := amount
f := make([]int, n+1)
for i := range f {
f[i] = 1 << 30
}
f[0] = 0
for _, x := range coins {
for j := x; j <= n; j++ {
f[j] = min(f[j], f[j-x]+1)
}
}
if f[n] > n {
return -1
}
return f[n]
}
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| function coinChange(coins: number[], amount: number): number {
const n = amount;
const f: number[] = Array(n + 1).fill(1 << 30);
f[0] = 0;
for (const x of coins) {
for (let j = x; j <= n; ++j) {
f[j] = Math.min(f[j], f[j - x] + 1);
}
}
return f[n] > n ? -1 : f[n];
}
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| /**
* @param {number[]} coins
* @param {number} amount
* @return {number}
*/
var coinChange = function (coins, amount) {
const n = amount;
const f = Array(n + 1).fill(1 << 30);
f[0] = 0;
for (const x of coins) {
for (let j = x; j <= n; ++j) {
f[j] = Math.min(f[j], f[j - x] + 1);
}
}
return f[n] > n ? -1 : f[n];
};
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