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# 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 Practice

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.

1. 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”.
2. 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.
3. Swap the above found two digits, we get 536974 in above example.
4. 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 ` `#include ` `#include ` `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); ` `    ``} ` `} `

## Python3

 `# 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 ``=` `list``(``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 `

## Javascript

 ``

Output

`Next number with same set of digits is 536479`

Time Complexity: O(N*logN)
Auxiliary Space: O(1)

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).

Optimised Approach:

1. Here instead of sorting the digits after (i-1) index, we are reversing the digits as mentioned in the above optimisation point.
2. As they will be in decreasing order so to find the smallest element possible from the right part we just reverse them thus reducing time complexity.

Below is the implementation of the above approach:

## C++14

 `#include ` `using` `namespace` `std; ` ` `  `vector<``int``> nextPermutation(``int` `n, vector<``int``> arr) ` `{ ` `    ``// If number of digits is 1 then just return the vector ` `    ``if` `(n == 1) ` `        ``return` `arr; ` ` `  `    ``// Start from the right most digit and find the first ` `    ``// digit that is ` `    ``// smaller than the digit next to it. ` `    ``int` `i = 0; ` `    ``for` `(i = n - 1; i > 0; i--) { ` `        ``if` `(arr[i] > arr[i - 1]) ` `            ``break``; ` `    ``} ` ` `  `    ``// If there is a possibility of a next greater element ` `    ``if` `(i != 0) { ` `        ``// Find the smallest digit on right side of (i-1)'th ` `        ``// digit that is ` `        ``// greater than number[i-1] ` `        ``for` `(``int` `j = n - 1; j >= i; j--) { ` `            ``if` `(arr[i - 1] < arr[j]) { ` `                ``// Swap the found smallest digit i.e. arr[j] ` `                ``// with arr[i-1] ` `                ``swap(arr[i - 1], arr[j]); ` `                ``break``; ` `            ``} ` `        ``} ` `    ``} ` ` `  `    ``// Reverse the digits after (i-1) because the digits ` `    ``// after (i-1) are in decreasing order and thus we will ` `    ``// get the smallest element possible from these digits ` `    ``reverse(arr.begin() + i, arr.end()); ` ` `  `    ``// If i is 0 that means elements are in decreasing order ` `    ``// Therefore, no greater element possible then we just ` `    ``// return the lowest possible ` `    ``// order/element formed from these digits by just ` `    ``// reversing the vector ` ` `  `    ``return` `arr; ` `} ` ` `  `int` `main() ` `{ ` `    ``int` `n = 6; ` `    ``vector<``int``> v{ 5,3,4,9,7,6 }; ` `    ``vector<``int``> res; ` `    ``res = nextPermutation(n, v); ` `    ``for` `(``int` `i = 0; i < res.size(); i++) { ` `        ``cout << res[i] << ``" "``; ` `    ``} ` `}`

## Java

 `// Java implementation of the approach ` `import` `java.util.*; ` ` `  `class` `GFG ` `{ ` `  ``static` `int``[] nextPermutation(``int` `n, ``int``[] arr) ` `  ``{ ` ` `  `    ``// If number of digits is 1 then just return the vector ` `    ``if` `(n == ``1``) ` `      ``return` `arr; ` ` `  `    ``// Start from the right most digit and find the first ` `    ``// digit that is ` `    ``// smaller than the digit next to it. ` `    ``int` `i = ``0``; ` `    ``for` `(i = n - ``1``; i > ``0``; i--) { ` `      ``if` `(arr[i] > arr[i - ``1``]) ` `        ``break``; ` `    ``} ` ` `  `    ``// If there is a possibility of a next greater element ` `    ``if` `(i != ``0``) ` `    ``{ ` ` `  `      ``// Find the smallest digit on right side of (i-1)'th ` `      ``// digit that is ` `      ``// greater than number[i-1] ` `      ``for` `(``int` `j = n - ``1``; j >= i; j--)  ` `      ``{ ` `        ``if` `(arr[i - ``1``] < arr[j]) ` `        ``{ ` ` `  `          ``// Swap the found smallest digit i.e. arr[j] ` `          ``// with arr[i-1] ` `          ``int` `temp = arr[j]; ` `          ``arr[j] = arr[i - ``1``]; ` `          ``arr[i - ``1``] = temp; ` `          ``break``; ` `        ``} ` `      ``} ` `    ``} ` ` `  `    ``// Reverse the digits after (i-1) because the digits ` `    ``// after (i-1) are in decreasing order and thus we will ` `    ``// get the smallest element possible from these digits ` `    ``int``[] res = ``new` `int``[arr.length]; ` `    ``int` `ind = arr.length - ``1``; ` ` `  `    ``// copying the first i elements of arr ` `    ``// into res ` `    ``for` `(``int` `j = ``0``; j < i; j++) ` `      ``res[j] = arr[j]; ` ` `  `    ``// copying the rest of the elements ` `    ``// in reverse order ` `    ``for` `(``int` `j = i; j< res.length; j++) ` `      ``res[j] = arr[ind--]; ` ` `  `    ``// If i is 0 that means elements are in decreasing order ` `    ``// Therefore, no greater element possible then we just ` `    ``// return the lowest possible ` `    ``// order/element formed from these digits by just ` `    ``// reversing the vector ` `    ``return` `res; ` `  ``} ` ` `  `  ``// Driver Code ` `  ``public` `static` `void` `main(String[] args) ` `  ``{ ` `    ``int` `n = ``6``; ` `    ``int``[] v = { ``5``,``3``,``4``,``9``,``7``,``6` `}; ` `    ``int``[] res; ` ` `  `    ``// Function Call ` `    ``res = nextPermutation(n, v); ` `    ``for` `(``int` `i = ``0``; i < res.length; i++) { ` `      ``System.out.print(res[i] + ``" "``); ` `    ``} ` `  ``} ` `} ` ` `  `// This code is contributed by phasing17`

## Python3

 `# A python program to find the next greatest number ` `def` `nextPermutation(arr): ` `   `  `      ``# find the length of the array ` `    ``n ``=` `len``(arr) ` `     `  `    ``# start from the right most digit and find the first ` `    ``# digit that is smaller than the digit next to it. ` `    ``k ``=` `n ``-` `2` `    ``while` `k >``=` `0``: ` `        ``if` `arr[k] < arr[k ``+` `1``]: ` `            ``break` `        ``k ``-``=` `1` `         `  `    ``# reverse the list if the digit that is smaller than the ` `    ``# digit next to it is not found. ` `    ``if` `k < ``0``: ` `        ``arr ``=` `arr[::``-``1``] ` `    ``else``: ` `       `  `          ``# find the first greatest element than arr[k] from the  ` `        ``# end of the list ` `        ``for` `l ``in` `range``(n ``-` `1``, k, ``-``1``): ` `            ``if` `arr[l] > arr[k]: ` `                ``break` ` `  `        ``# swap the elements at arr[k] and arr[l       ` `        ``arr[l], arr[k] ``=` `arr[k], arr[l] ` `         `  `        ``# reverse the list from k + 1 to the end to find the  ` `        ``# most nearest greater number to the given input number ` `        ``arr[k ``+` `1``:] ``=` `reversed``(arr[k ``+` `1``:]) ` ` `  `    ``return` `arr ` ` `  `# Driver code ` `arr ``=` `[``5``, ``3``, ``4``, ``9``, ``7``, ``6``] ` `print``(``*``nextPermutation(arr)) ` ` `  `# This code is contributed by Manish Thapa `

## C#

 `// C# implementation of the approach ` `using` `System; ` `class` `GFG { ` `  ``static` `int``[] nextPermutation(``int` `n, ``int``[] arr) ` `  ``{ ` ` `  `    ``// If number of digits is 1 then just return the ` `    ``// vector ` `    ``if` `(n == 1) ` `      ``return` `arr; ` ` `  `    ``// Start from the right most digit and find the ` `    ``// first digit that is smaller than the digit next ` `    ``// to it. ` `    ``int` `i = 0; ` `    ``for` `(i = n - 1; i > 0; i--) { ` `      ``if` `(arr[i] > arr[i - 1]) ` `        ``break``; ` `    ``} ` ` `  `    ``// If there is a possibility of a next greater ` `    ``// element ` `    ``if` `(i != 0) { ` ` `  `      ``// Find the smallest digit on right side of ` `      ``// (i-1)'th digit that is greater than ` `      ``// number[i-1] ` `      ``for` `(``int` `j = n - 1; j >= i; j--) { ` `        ``if` `(arr[i - 1] < arr[j]) { ` ` `  `          ``// Swap the found smallest digit i.e. ` `          ``// arr[j] with arr[i-1] ` `          ``int` `temp = arr[j]; ` `          ``arr[j] = arr[i - 1]; ` `          ``arr[i - 1] = temp; ` `          ``break``; ` `        ``} ` `      ``} ` `    ``} ` ` `  `    ``// Reverse the digits after (i-1) because the digits ` `    ``// after (i-1) are in decreasing order and thus we ` `    ``// will get the smallest element possible from these ` `    ``// digits ` `    ``int``[] res = ``new` `int``[arr.Length]; ` `    ``int` `ind = arr.Length - 1; ` ` `  `    ``// copying the first i elements of arr ` `    ``// into res ` `    ``for` `(``int` `j = 0; j < i; j++) ` `      ``res[j] = arr[j]; ` ` `  `    ``// copying the rest of the elements ` `    ``// in reverse order ` `    ``for` `(``int` `j = i; j < res.Length; j++) ` `      ``res[j] = arr[ind--]; ` ` `  `    ``// If i is 0 that means elements are in decreasing ` `    ``// order Therefore, no greater element possible then ` `    ``// we just return the lowest possible order/element ` `    ``// formed from these digits by just reversing the ` `    ``// vector ` `    ``return` `res; ` `  ``} ` ` `  `  ``// Driver Code ` `  ``public` `static` `int` `Main() ` `  ``{ ` `    ``int` `n = 6; ` `    ``int``[] v = ``new` `int``[] { 5, 3, 4, 9, 7, 6 }; ` `    ``int``[] res; ` ` `  `    ``// Function Call ` `    ``res = nextPermutation(n, v); ` `    ``for` `(``int` `i = 0; i < res.Length; i++) { ` `      ``Console.Write(res[i] + ``" "``); ` `    ``} ` `    ``return` `0; ` `  ``} ` `} ` ` `  `// This code is contributed by Taranpreet`

## Javascript

 ` `

Output

`5 3 6 4 7 9 `

Time Complexity: O(N)
Auxiliary Space: O(1)

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