Form the Smallest Number Using At Most One Swap Operation in C++

In this problem, we are given a positive integer. Our task is to create a program to form the smaller number using at most one swap operation.

We will be creating a new number using the digits of the existing number. The smallest number formed can have only one digit swapped from the existing number.

Let’s take an example to understand the problem

Input: n = 63519
Output: 36519

Solution Approach

One method to solve the problem is by finding all the numbers created by swapping pair of digits of the given number. Out of all these swapped digit numbers, the smallest one is returned. For this, we will convert the number to string and swap positions.

Example

Program to illustrate the working of our solution

#include <iostream>
using namespace std;

int findSmallestNumSwapDig(int N){

   string strNum = to_string(N);
   string temp = strNum;
   for (int i = 0; i < strNum.size(); i++) {
      for (int j = i + 1; j < strNum.size(); j++) {
         swap(strNum[i], strNum[j]);
         if (stoi(strNum) < stoi(temp))
            temp = strNum;
         swap(strNum[i], strNum[j]);
      }
   }
   return stoi(temp);
}
int main(){
   int num = 792156;
   cout<<"The number is "<<num<<endl;
   cout<<"The smallest number created by swapping one digit is "<<findSmallestNumSwapDig(num) << endl;
   return 0;
}

Output

The number is 792156
The smallest number created by swapping one digit is192756

Another approach

One more approach to solve the problem is by using an extra auxiliary array aux[]. This array is used to store the index of the smallest digit at the right (greater index) of the current index i.e. aux[i] is the index of the smallest digit at the right side of the number’s digit. Initialise aux[i] = -1, if no such digit exists. Now, traverse the number from index 0 to n-1 and find the number in the aux[] array which is smaller than the current value. For index 0 value, check for non-zero element, if 0 occurs discard the value. If any element is found, swap arr[i] & arr[aux[i]] and return the created number.

Example

Program to illustrate the working of our solution

#include <bits/stdc++.h>
using namespace std;

int findSmallestNumSwapDig(int N){

   string num = to_string(N);
   int n = num.size();
   int auxArr[n], right;
   auxArr[n - 1] = -1;
   right = n - 1;
   for (int i = n - 2; i >= 1; i--) {
      if (num[i] >= num[right])
         auxArr[i] = right;
      else {
         if (num[i] == num[i + 1])
            auxArr[i] = right;
         else {
            auxArr[i] = -1;
            right = i;
         }
      }
   }
   int small = -1;
   for (int i = 1; i < n; i++)
   if (num[i] != '0') {
      if (small == -1) {
         if (num[i] < num[0])
            small = i;
      }
      else if (num[i] <= num[small])
            small = i;
   }
   if (small != -1)
      swap(num[0], num[small]);
   else {
      for (int i = 1; i < n; i++) {
         if (auxArr[i] != -1 && num[i] != num[auxArr[i]]) {
            swap(num[i], num[auxArr[i]]);
            break;
         }
      }
   }
   return stoi(num);
}
int main(){
   int num = 792156;
   cout<<"The number is "<<num<<endl;
   cout<<"The smallest number created by swapping one digit is" <<findSmallestNumSwapDig(num)<< endl;
      return 0;
}

Output

The number is 792156
The smallest number created by swapping one digit is192756