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Largest Multiple of Three in C++
Suppose we have one array of digits, we have to find the largest multiple of three that can be formed by concatenating some of the given digits in any order as we want. The answer may be very large so make it as string. If there is no answer return an empty string.
So, if the input is like [7,2,8], then the output will be 87
To solve this, we will follow these steps −
Define one 2D array d, there will be three rows
sort the array digits
sum := 0
for initialize i := 0, when i < size of digits, update (increase i by 1), do−
x := digits[i]
insert digits[i] at the end of d[x mod 3]
sum := sum + x
sum := sum mod 3
if sum is non-zero, then −
if not size of d[sum], then −
rem := 3 - sum
if size of d[rem] < 2, then −
return empty string
delete last element from d[rem] twice
Otherwise
delete last element from d[sum]
ret := empty string
for initialize i := 0, when i < 3, update (increase i by 1), do −
for initialize j := 0, when j < size of d[i], update (increase j by 1), do−
ret := ret concatenate d[i, j] as string
sort the array ret
if size of ret and ret[0] is same as '0', then −
return "0"
return "0"
Let us see the following implementation to get better understanding −
Example
#include <bits/stdc++.h>
using namespace std;
class Solution {
public:
string largestMultipleOfThree(vector<int>& digits) {
vector<vector<int>> d(3);
sort(digits.begin(), digits.end(), greater<int>());
int sum = 0;
for (int i = 0; i < digits.size(); i++) {
int x = digits[i];
d[x % 3].push_back(digits[i]);
sum += x;
sum %= 3;
}
if (sum) {
if (!d[sum].size()) {
int rem = 3 - sum;
if (d[rem].size() < 2)
return "";
d[rem].pop_back();
d[rem].pop_back();
}
else {
d[sum].pop_back();
}
}
string ret = "";
for (int i = 0; i < 3; i++) {
for (int j = 0; j < d[i].size(); j++) {
ret += to_string(d[i][j]);
}
}
sort(ret.begin(), ret.end(), greater<int>());
if (ret.size() && ret[0] == '0')
return "0";
return ret;
}
};
main(){
Solution ob;
vector<int> v = {7,2,8};
cout << (ob.largestMultipleOfThree(v));
}Input
{7,2,8}Output
87
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