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.


0 ≤ n < 2 ** 31

Example 1


n = 1929




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]);

        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

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