# Form the largest number using at most one swap operation

Given a non-negative number num. The problem is to apply at most one swap operation on the number num so that the resultant is the largest possible number. The number could be very large so a string type can be used to store the number.

Examples:

```Input : n = 8725634
Output : 8765234
Swapped the digits 2 and 6.

Input : n = 54321
Output : 54321
No swapping of digits required.
```

## Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

Create an array rightMax[]. rightMax[i] contains the index of the greatest digit which is on the right side of num[i] and also greater than num[i]. If no such digit exists then rightMax[i] = -1. Now, traverse the rightMax[] array from i = 0 to n-1(where n is the total number of digits in num), and find the first element having rightMax[i] != -1. Perform the swap(num[i], num[rightMax[i]]) operation and break.

## C++

```// C++ implementation to form the largest number
// by applying atmost one swap operation
#include <bits/stdc++.h>
using namespace std;

// function to form the largest number by
// applying atmost one swap operation
string largestNumber(string num)
{
int n = num.size();
int rightMax[n], right;

// for the rightmost digit, there
// will be no greater right digit
rightMax[n - 1] = -1;

// index of the greatest right digit till the
// current index from the right direction
right = n - 1;

// traverse the array from second right element
// up to the left element
for (int i = n - 2; i >= 0; i--) {

// if 'num[i]' is less than the greatest digit
// encountered so far
if (num[i] < num[right])
rightMax[i] = right;

// else
else {
// there is no greater right digit
// for 'num[i]'
rightMax[i] = -1;

// update 'right' index
right = i;
}
}

// traverse the 'rightMax[]' array from left to right
for (int i = 0; i < n; i++) {

// if for the current digit, greater right digit exists
// then swap it with its greater right digit and break
if (rightMax[i] != -1) {

// performing the required swap operation
swap(num[i], num[rightMax[i]]);
break;
}
}

// required largest number
return num;
}

// Driver program to test above
int main()
{
string num = "8725634";
cout << "Largest number:"
<< largestNumber(num);
return 0;
}
```

## Java

```//Java implementation to form the largest number
//by applying utmost one swap operation
public class GFG
{
// function to form the largest number by
// applying utmost one swap operation
static String largestNumber(String num)
{
int n = num.length();
int right;
int rightMax[] = new int[n];

// for the rightmost digit, there
// will be no greater right digit
rightMax[n - 1] = -1;

// index of the greatest right digit
// till the current index from the
// right direction
right = n - 1;

// traverse the array from second right
// element up to the left element
for (int i = n - 1; i >= 0 ; i--)
{
// if 'num.charAt(i)' is less than the
// greatest digit encountered so far
if (num.charAt(i) < num.charAt(right))
rightMax[i] = right;

else
{
// there is no greater right digit
// for 'num.charAt(i)'
rightMax[i] = -1;

// update 'right' index
right = i;
}
}

// traverse the 'rightMax[]' array from
// left to right
for (int i = 0; i < n; i++)
{

// if for the current digit, greater
// right digit exists then swap it
// with its greater right digit and break
if (rightMax[i] != -1)
{
// performing the required swap operation
num = swap(num,i,rightMax[i]);
break;
}
}

// required largest number
return num;
}

// Utility method to swap two characters
// in a String
static String swap(String num, int i, int j)
{
StringBuilder sb= new StringBuilder(num);
sb.setCharAt(i, num.charAt(j));
sb.setCharAt(j, num.charAt(i));
return sb.toString();

}

//Driver Function to test above Function
public static void main(String[] args)
{
String num = "8725634";
System.out.println("Largest Number : " +
largestNumber(num));
}

}
//This code is contributed by Sumit Ghosh
```

## Python

```# Python implementation to form the largest number
# by applying atmost one swap operation

# function to form the largest number by
# applying atmost one swap operation
def largestNumber(num):
n = len(num)
rightMax = [0 for i in range(n)]

# for the rightmost digit, there
# will be no greater right digit
rightMax[n - 1] = -1

# index of the greatest right digit till the
# current index from the right direction
right = n - 1

# traverse the array from second right element
# up to the left element
i = n - 2
while i >= 0:

# if 'num[i]' is less than the greatest digit
# encountered so far
if (num[i] < num[right]):
rightMax[i] = right

# else
else:
# there is no greater right digit
# for 'num[i]'
rightMax[i] = -1

# update 'right' index
right = i
i -= 1

# traverse the 'rightMax[]' array from left to right
for i in range(n):

# if for the current digit, greater right digit exists
# then swap it with its greater right digit and break
if (rightMax[i] != -1):

# performing the required swap operation
t = num[i]

num[i] = num[rightMax[i]]
num[rightMax[i]] = t
break

# required largest number
return num

# Driver program to test above
num = "8725634"
li = [i for i in num]
print "Largest number: "
li = largestNumber(li)
for i in li:
print i,
print

#This code is contributed by Sachin Bisht
```

Output:

```Largest number: 8765234
```

Time Complexity: O(n), where n is the total number of digits.
Auxiliary Space: O(n), where n is the total number of digits.

This article is contributed by Ayush Jauhari. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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