Description#
Given an array of digits digits, return the largest multiple of three that can be formed by concatenating some of the given digits in any order. If there is no answer return an empty string.
Since the answer may not fit in an integer data type, return the answer as a string. Note that the returning answer must not contain unnecessary leading zeros.
Example 1:
Input: digits = [8,1,9]
Output: "981"
Example 2:
Input: digits = [8,6,7,1,0]
Output: "8760"
Example 3:
Input: digits = [1]
Output: ""
Constraints:
1 <= digits.length <= 1040 <= digits[i] <= 9
Solutions#
Solution 1: Greedy + Dynamic Programming + Backtracking#
We define $f[i][j]$ as the maximum length of selecting several numbers from the first $i$ numbers, so that the sum of the selected numbers modulo $3$ equals $j$. To make the selected numbers as large as possible, we need to select as many numbers as possible, so we need to make $f[i][j]$ as large as possible. We initialize $f[0][0] = 0$, and the rest of $f[0][j] = -\infty$.
Consider how $f[i][j]$ transitions. We can choose not to select the $i$-th number, in which case $f[i][j] = f[i - 1][j]$; we can also choose to select the $i$-th number, in which case $f[i][j] = f[i - 1][(j - x_i \bmod 3 + 3) \bmod 3] + 1$, where $x_i$ represents the value of the $i$-th number. Therefore, we have the following state transition equation:
$$
f[i][j] = \max { f[i - 1][j], f[i - 1][(j - x_i \bmod 3 + 3) \bmod 3] + 1 }
$$
If $f[n][0] \le 0$, then we cannot select any number, so the answer string is empty. Otherwise, we can backtrack through the $f$ array to find out the selected numbers.
Define $i = n$, $j = 0$, start backtracking from $f[i][j]$, let $k = (j - x_i \bmod 3 + 3) \bmod 3$, if $f[i - 1][k] + 1 = f[i][j]$, then we have selected the $i$-th number, otherwise we have not selected the $i$-th number. If we have selected the $i$-th number, then we update $j$ to $k$, otherwise we keep $j$ unchanged. To make the number of the same length as large as possible, we should prefer to select larger numbers, so we should sort the array first.
The time complexity is $O(n \times \log n)$, and the space complexity is $O(n)$. Where $n$ is the length of the array.
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| class Solution:
def largestMultipleOfThree(self, digits: List[int]) -> str:
digits.sort()
n = len(digits)
f = [[-inf] * 3 for _ in range(n + 1)]
f[0][0] = 0
for i, x in enumerate(digits, 1):
for j in range(3):
f[i][j] = max(f[i - 1][j], f[i - 1][(j - x % 3 + 3) % 3] + 1)
if f[n][0] <= 0:
return ""
arr = []
j = 0
for i in range(n, 0, -1):
k = (j - digits[i - 1] % 3 + 3) % 3
if f[i - 1][k] + 1 == f[i][j]:
arr.append(digits[i - 1])
j = k
i = 0
while i < len(arr) - 1 and arr[i] == 0:
i += 1
return "".join(map(str, arr[i:]))
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| class Solution {
public String largestMultipleOfThree(int[] digits) {
Arrays.sort(digits);
int n = digits.length;
int[][] f = new int[n + 1][3];
final int inf = 1 << 30;
for (var g : f) {
Arrays.fill(g, -inf);
}
f[0][0] = 0;
for (int i = 1; i <= n; ++i) {
for (int j = 0; j < 3; ++j) {
f[i][j] = Math.max(f[i - 1][j], f[i - 1][(j - digits[i - 1] % 3 + 3) % 3] + 1);
}
}
if (f[n][0] <= 0) {
return "";
}
StringBuilder sb = new StringBuilder();
for (int i = n, j = 0; i > 0; --i) {
int k = (j - digits[i - 1] % 3 + 3) % 3;
if (f[i - 1][k] + 1 == f[i][j]) {
sb.append(digits[i - 1]);
j = k;
}
}
int i = 0;
while (i < sb.length() - 1 && sb.charAt(i) == '0') {
++i;
}
return sb.substring(i);
}
}
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| class Solution {
public:
string largestMultipleOfThree(vector<int>& digits) {
sort(digits.begin(), digits.end());
int n = digits.size();
int f[n + 1][3];
memset(f, -0x3f, sizeof(f));
f[0][0] = 0;
for (int i = 1; i <= n; ++i) {
for (int j = 0; j < 3; ++j) {
f[i][j] = max(f[i - 1][j], f[i - 1][(j - digits[i - 1] % 3 + 3) % 3] + 1);
}
}
if (f[n][0] <= 0) {
return "";
}
string ans;
for (int i = n, j = 0; i; --i) {
int k = (j - digits[i - 1] % 3 + 3) % 3;
if (f[i - 1][k] + 1 == f[i][j]) {
ans += digits[i - 1] + '0';
j = k;
}
}
int i = 0;
while (i < ans.size() - 1 && ans[i] == '0') {
++i;
}
return ans.substr(i);
}
};
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| func largestMultipleOfThree(digits []int) string {
sort.Ints(digits)
n := len(digits)
const inf = 1 << 30
f := make([][]int, n+1)
for i := range f {
f[i] = make([]int, 3)
for j := range f[i] {
f[i][j] = -inf
}
}
f[0][0] = 0
for i := 1; i <= n; i++ {
for j := 0; j < 3; j++ {
f[i][j] = max(f[i-1][j], f[i-1][(j-digits[i-1]%3+3)%3]+1)
}
}
if f[n][0] <= 0 {
return ""
}
ans := []byte{}
for i, j := n, 0; i > 0; i-- {
k := (j - digits[i-1]%3 + 3) % 3
if f[i][j] == f[i-1][k]+1 {
ans = append(ans, byte('0'+digits[i-1]))
j = k
}
}
i := 0
for i < len(ans)-1 && ans[i] == '0' {
i++
}
return string(ans[i:])
}
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| function largestMultipleOfThree(digits: number[]): string {
digits.sort((a, b) => a - b);
const n = digits.length;
const f: number[][] = new Array(n + 1).fill(0).map(() => new Array(3).fill(-Infinity));
f[0][0] = 0;
for (let i = 1; i <= n; ++i) {
for (let j = 0; j < 3; ++j) {
f[i][j] = Math.max(f[i - 1][j], f[i - 1][(j - (digits[i - 1] % 3) + 3) % 3] + 1);
}
}
if (f[n][0] <= 0) {
return '';
}
const arr: number[] = [];
for (let i = n, j = 0; i; --i) {
const k = (j - (digits[i - 1] % 3) + 3) % 3;
if (f[i - 1][k] + 1 === f[i][j]) {
arr.push(digits[i - 1]);
j = k;
}
}
let i = 0;
while (i < arr.length - 1 && arr[i] === 0) {
++i;
}
return arr.slice(i).join('');
}
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