Find next greater number with same set of digits

Given a number n, find the smallest number that has same set of digits as n and is greater than n. If n is the greatest possible number with its set of digits, then print “not possible”.

Examples:
For simplicity of implementation, we have considered input number as a string.

Input:  n = "218765"
Output: "251678"

Input:  n = "1234"
Output: "1243"

Input: n = "4321"
Output: "Not Possible"

Input: n = "534976"
Output: "536479"

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

Following are few observations about the next greater number.
1) If all digits sorted in descending order, then output is always “Not Possible”. For example, 4321.
2) If all digits are sorted in ascending order, then we need to swap last two digits. For example, 1234.
3) For other cases, we need to process the number from rightmost side (why? because we need to find the smallest of all greater numbers)

You can now try developing an algorithm yourself.

Following is the algorithm for finding the next greater number.
I) Traverse the given number from rightmost digit, keep traversing till you find a digit which is smaller than the previously traversed digit. For example, if the input number is “534976”, we stop at 4 because 4 is smaller than next digit 9. If we do not find such a digit, then output is “Not Possible”.

II) Now search the right side of above found digit ‘d’ for the smallest digit greater than ‘d’. For “534976″, the right side of 4 contains “976”. The smallest digit greater than 4 is 6.

III) Swap the above found two digits, we get 536974 in above example.

IV) Now sort all digits from position next to ‘d’ to the end of number. The number that we get after sorting is the output. For above example, we sort digits in bold 536974. We get “536479” which is the next greater number for input 534976.

Following are the implementation of above approach.

C

// C++ program to find the smallest number which greater than a given number 
// and has same set of digits as given number 
#include <iostream> 
#include <cstring> 
#include <algorithm> 
using namespace std; 
  
// Utility function to swap two digits 
void swap(char *a, char *b) 
{ 
    char temp = *a; 
    *a = *b; 
    *b = temp; 
} 
  
// Given a number as a char array number[], this function finds the 
// next greater number.  It modifies the same array to store the result 
void findNext(char number[], int n) 
{ 
    int i, j; 
  
    // I) Start from the right most digit and find the first digit that is 
    // smaller than the digit next to it. 
    for (i = n-1; i > 0; i--) 
        if (number[i] > number[i-1]) 
           break; 
  
    // If no such digit is found, then all digits are in descending order 
    // means there cannot be a greater number with same set of digits 
    if (i==0) 
    { 
        cout << "Next number is not possible"; 
        return; 
    } 
  
    // II) Find the smallest digit on right side of (i-1)'th digit that is 
    // greater than number[i-1] 
    int x = number[i-1], smallest = i; 
    for (j = i+1; j < n; j++) 
        if (number[j] > x && number[j] < number[smallest]) 
            smallest = j; 
  
    // III) Swap the above found smallest digit with number[i-1] 
    swap(&number;[smallest], &number;[i-1]); 
  
    // IV) Sort the digits after (i-1) in ascending order 
    sort(number + i, number + n); 
  
    cout << "Next number with same set of digits is " << number; 
  
    return; 
} 
  
// Driver program to test above function 
int main() 
{ 
    char digits[] = "534976"; 
    int n = strlen(digits); 
    findNext(digits, n); 
    return 0; 
} 

Java

// Java program to find next greater  
// number with same set of digits. 
import java.util.Arrays; 
  
public class nextGreater  
{ 
    // Utility function to swap two digit 
    static void swap(char ar[], int i, int j)  
    { 
        char temp = ar[i]; 
        ar[i] = ar[j]; 
        ar[j] = temp; 
    } 
  
    // Given a number as a char array number[],  
    // this function finds the next greater number.  
    // It modifies the same array to store the result 
    static void findNext(char ar[], int n)  
    { 
        int i; 
          
        // I) Start from the right most digit  
        // and find the first digit that is smaller  
        // than the digit next to it. 
        for (i = n - 1; i > 0; i--)  
        { 
            if (ar[i] > ar[i - 1]) { 
                break; 
            } 
        } 
          
        // If no such digit is found, then all  
        // digits are in descending order means  
        // there cannot be a greater number with  
        // same set of digits 
        if (i == 0)  
        { 
            System.out.println("Not possible"); 
        }  
        else 
        { 
            int x = ar[i - 1], min = i; 
              
            // II) Find the smallest digit on right  
            // side of (i-1)'th digit that is greater  
            // than number[i-1] 
            for (int j = i + 1; j < n; j++)  
            { 
                if (ar[j] > x && ar[j] < ar[min])  
                { 
                    min = j; 
                } 
            } 
  
            // III) Swap the above found smallest  
            // digit with number[i-1] 
            swap(ar, i - 1, min); 
  
            // IV) Sort the digits after (i-1)  
            // in ascending order 
            Arrays.sort(ar, i, n); 
            System.out.print("Next number with same" + 
                                    " set of digits is "); 
            for (i = 0; i < n; i++) 
                System.out.print(ar[i]); 
        } 
    } 
  
    public static void main(String[] args)  
    { 
        char digits[] = { '5','3','4','9','7','6' }; 
        int n = digits.length; 
        findNext(digits, n); 
    } 
} 

Python

# Python program to find the smallest number which  
# is greater than a given no. has same set of  
# digits as given number 
  
# Given number as int array, this function finds the  
# greatest number and returns the number as integer 
def findNext(number,n): 
       
     # Start from the right most digit and find the first 
     # digit that is smaller than the digit next to it 
     for i in range(n-1,0,-1): 
         if number[i] > number[i-1]: 
             break
               
     # If no such digit found,then all numbers are in  
     # descending order, no greater number is possible 
     if i == 1 and number[i] <= number[i-1]: 
         print "Next number not possible"
         return
           
     # Find the smallest digit on the right side of  
     # (i-1)'th digit that is greater than number[i-1] 
     x = number[i-1] 
     smallest = i 
     for j in range(i+1,n): 
         if number[j] > x and number[j] < number[smallest]: 
             smallest = j 
           
     # Swapping the above found smallest digit with (i-1)'th 
     number[smallest],number[i-1] = number[i-1], number[smallest] 
       
     # X is the final number, in integer datatype  
     x = 0
     # Converting list upto i-1 into number 
     for j in range(i): 
         x = x * 10 + number[j] 
       
     # Sort the digits after i-1 in ascending order 
     number = sorted(number[i:]) 
     # converting the remaining sorted digits into number 
     for j in range(n-i): 
         x = x * 10 + number[j] 
       
     print "Next number with set of digits is",x 
  
  
# Driver Program to test above function 
digits = "534976"         
  
# converting into integer array, 
# number becomes [5,3,4,9,7,6] 
number = map(int ,digits)  
findNext(number, len(digits)) 
  
# This code is contributed by Harshit Agrawal 

C#

// C# program to find next greater  
// number with same set of digits. 
using System; 
                      
public class nextGreater  
{ 
    // Utility function to swap two digit 
    static void swap(char []ar, int i, int j)  
    { 
        char temp = ar[i]; 
        ar[i] = ar[j]; 
        ar[j] = temp; 
    } 
  
    // Given a number as a char array number[],  
    // this function finds the next greater number.  
    // It modifies the same array to store the result 
    static void findNext(char []ar, int n)  
    { 
        int i; 
          
        // I) Start from the right most digit  
        // and find the first digit that is smaller  
        // than the digit next to it. 
        for (i = n - 1; i > 0; i--)  
        { 
            if (ar[i] > ar[i - 1])  
            { 
                break; 
            } 
        } 
          
        // If no such digit is found, then all  
        // digits are in descending order means  
        // there cannot be a greater number with  
        // same set of digits 
        if (i == 0)  
        { 
            Console.WriteLine("Not possible"); 
        }  
        else
        { 
            int x = ar[i - 1], min = i; 
              
            // II) Find the smallest digit on right  
            // side of (i-1)'th digit that is greater  
            // than number[i-1] 
            for (int j = i + 1; j < n; j++)  
            { 
                if (ar[j] > x && ar[j] < ar[min])  
                { 
                    min = j; 
                } 
            } 
  
            // III) Swap the above found smallest  
            // digit with number[i-1] 
            swap(ar, i - 1, min); 
  
            // IV) Sort the digits after (i-1)  
            // in ascending order 
            Array.Sort(ar, i, n-i); 
            Console.Write("Next number with same" + 
                                    " set of digits is "); 
            for (i = 0; i < n; i++) 
                Console.Write(ar[i]); 
        } 
    } 
  
    // Driver code 
    public static void Main(String[] args)  
    { 
        char []digits = { '5','3','4','9','7','6' }; 
        int n = digits.Length; 
        findNext(digits, n); 
    } 
} 
  
// This code is contributed by 29AjayKumar 

Output:

Next number with same set of digits is 536479

The above implementation can be optimized in following ways.
1) We can use binary search in step II instead of linear search.
2) In step IV, instead of doing simple sort, we can apply some clever technique to do it in linear time. Hint: We know that all digits are linearly sorted in reverse order except one digit which was swapped.

With above optimizations, we can say that the time complexity of this method is O(n).

This article is contributed by Rahul Jain. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above

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My Personal Notes

Nov	DEC	Jan
	22
2019	2020	2021

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

C

Java

Python

C#

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