Find next greater number with same set of digits - GeeksforGeeks

14 Jan, 2022
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"
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++
Java
Python3
C#
Javascript
// 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 digitsvoid 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 resultvoid 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 functionint main(){    char digits[] = "534976";    int n = strlen(digits);    findNext(digits, n);    return 0;}
// 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 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 integerdef 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 functiondigits = "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# 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
<script> // JavaScript 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 integerfunction findNext(number, n){         // Start from the right most digit    // and find the first digit that is    // smaller than the digit next to it    for(var i = n - 1; i >= 0; i--)    {        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 && number[i] <= number[i - 1])    {        document.write("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]    let x = number[i - 1];    let smallest = i;         for(let j = i + 1; j < n; j++)    {        if (number[j] > x &&            number[j] < number[smallest])        smallest = j;    }         // Swapping the above found smallest    // digit with (i-1)'th    let temp = number[smallest];    number[smallest] = number[i - 1];    number[i - 1] = temp;         // X is the final number, in integer datatype    x = 0         // Converting list upto i-1 into number    for(let j = 0; j < i; j++)        x = x * 10 + number[j];         // Sort the digits after i-1 in ascending order    number = number.slice(i, number.length + 1);    number.sort()         // Converting the remaining sorted    // digits into number    for(let j = 0; j < n - i; j++)        x = x * 10 + number[j];         document.write("Next number with " +                   "set of digits is " + x);} // Driver codelet digits = "534976"       // Converting into integer array,// number becomes [5,3,4,9,7,6]let number = []for(let i = 0; i < digits.length; i++)    number[i] = Number(digits[i]); findNext(number, digits.length); // This code is contributed by rohan07 </script>
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
Python3
Javascript
#include <bits/stdc++.h>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] << " ";    }}
# A python program to find the next greatest numberdef 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 codearr = [5, 3, 4, 9, 7, 6]print(*nextPermutation(arr)) # This code is contributed by Manish Thapa
<script> function nextPermutation(n, 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.    let 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 (let 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]                                 let temp = arr[i - 1];                arr[i - 1] = arr[j];                arr[j] = 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    arr = arr.slice(0,i).concat(arr.slice(i,arr.length).reverse());      // 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;} let v = [5,3,4,9,7,6];let n = 6;let res = nextPermutation(n, v);for (let i = 0; i < res.length; i++) {    document.write(res[i] + " ")} // This code is contributed by avanitrachhadiya2155</script>
5 3 6 4 7 9 
Time Complexity: O(N)Auxiliary Space: O(1)
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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Morgan Stanley
Amazon
Microsoft
Samsung
Snapdeal
FactSet
Hike
MakeMyTrip
Adobe
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Oxigen Wallet
Vizury Interactive Solutions
Arrays
Searching
Mathematical
permutation
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