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Java-based LeetCode algorithm problem solutions, regularly updated
package g1301_1400.s1363_largest_multiple_of_three;
// #Hard #Array #Dynamic_Programming #Greedy #2022_03_22_Time_5_ms_(84.67%)_Space_53.3_MB_(70.80%)
import java.util.Arrays;
public class Solution {
public String largestMultipleOfThree(int[] digits) {
int sum = 0;
int[] count = new int[10];
// Here we are using the property that any no is divisible by 3 when its sum of digits is
// divisible by 3
// get sum of digits and count of each digit
for (int x : digits) {
sum += x;
count[x]++;
}
StringBuilder sb = new StringBuilder();
int[] copied = Arrays.copyOf(count, count.length);
// if sum % 3 != 0 then processing required
if (sum % 3 != 0) {
int rem = sum % 3;
int oldRem = rem;
while (oldRem != 0) {
while (rem != 0) {
// if the remainder that we are trying to delete and its required digits is not
// present
// then the value will become -ve at that digit
copied[rem % 10]--;
// increase the remainder by 3 each time a -ve value is found
// and reset the rem and copied from orig count array and break
if (copied[rem % 10] < 0) {
oldRem += 3;
rem = oldRem;
copied = Arrays.copyOf(count, count.length);
break;
}
rem /= 10;
if (rem == 0) {
oldRem = 0;
}
}
}
}
// generate the largest number by considering from the last digit ie 9,8,7,6...
for (int i = 9; i >= 0; i--) {
int val = copied[i];
while (val > 0) {
sb.append(i);
val--;
}
}
// check for any leading zeroes and remove
while (sb.length() > 1) {
if (sb.charAt(0) != '0') {
break;
} else {
sb.deleteCharAt(0);
}
}
return sb.toString();
}
}