# Maximum Number After One Swap - Facebook Top Interview Questions

### Problem Statement :

```Given an integer n, you can swap any two digits at most once.

Return the maximum value of the resulting number.

Constraints

0 ≤ n < 2 ** 31

Example 1

Input

n = 1929

Output

9921

Explanation

We swap the first and the last digits```

### Solution :

```                        ```Solution in C++ :

int solve(int n) {
if (n == 0) return 0;

string s = to_string(n);
int len = s.size();

vector<deque<int> > indices(10);

for (int i = 0; i < len; i++) {
indices[s[i] - '0'].push_back(i);
}
bool found = false;
for (int i = 0; i < len; i++) {
for (int j = 9; j > s[i] - '0'; j--) {
if (indices[j].size() > 0) {
int last = indices[j].back();
found = true;
swap(s[i], s[last]);
break;
}
}

if (found == true) break;
indices[s[i] - '0'].pop_front();
}

return stoi(s);
}```
```

```                        ```Solution in Java :

import java.util.*;

class Solution {
public int solve(int n) {
char[] ch = String.valueOf(n).toCharArray();
for (int i = 0; i < ch.length - 1; i++) {
int max = -1, ind = -1;
for (int j = i + 1; j < ch.length; j++) {
if (ch[j] - '0' >= max) {
max = ch[j] - '0';
ind = j;
}
}
if (max > ch[i] - '0') {
char temp = ch[i];
ch[i] = ch[ind];
ch[ind] = temp;
return Integer.parseInt(new String(ch));
}
}
return n;
}
}```
```

```                        ```Solution in Python :

class Solution:
def solve(self, n):
st = list(str(n))
curmax = n
for i in range(len(st)):
for j in range(i + 1, len(st)):
if st[j] > st[i]:
st[i], st[j] = st[j], st[i]
curmax = max(curmax, int("".join(st)))
st[i], st[j] = st[j], st[i]
return curmax```
```

## Print in Reverse

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## Get Node Value

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