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Next Greater Same Digit

Last Updated on March 25, 2022 by Ria Pathak

Searching

Hard

PROBLEM STATEMENT`(`SIMPLIFIED`)`:

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 the same number `N`.

See original problem statement here

For Example:

``````Input  : "12345"

Output : "12354"``````
``````Input  : "54321"

Output : "54321"

Explanation : No number is greater than 54321 with same set of digits, thats why print same number.``````
``````Input  : "12321"

Output : "13122"``````

OBSERVATIONS:

1. If the digits are all sorted in strictly descending order, then the number itself would be our answer as there is no bigger number than that.
For example, `54321`.
2. If the digits are all sorted in strictly increasing order, then just swap the last two digits. For example, `12345` will change to `12354`.
3. For all other cases, we need to process all the digits in the number from right to left as we need to find the smallest of the greater numbers.

SOLVING APPROACH:

1. Start traversing the given number from rightmost digit till you reach a digit that is just smaller than its right digit, say index of this digit be `i`.
For example:
`N` `=` `125321`, here if we scan from right to left, `2` is the first digit that is less than `5` so `2` is our `ith` index digit.
2. Now again traverse to the right of this index `i` and find the smallest digit that is just greater than the digit at `ith` index. Let’s say the found index is `j` at the right of `i`. Now if we traverse right of `2`, the smallest digit that is greater than `2` is `3` (not `5`).
3. Swap digits at `ith` and `jth` index. After swapping `125321` becomes `135221`.
4. Sort digits from `(i+1)` to the rightmost index in increasing order. After sorting, `135221` becomes `131225`, which is our required solution.

OPTIMIZATIONS:

1. A few optimizations can be made in this approach and significantly improve the `Time Complexity` of the approach.
2. The `Linear Search` used in `Step-2` can be replaced with `Binary Search` as the right digits are already sorted. This reduces `Time Complexity` from `O(n)` to `O(logn)`.
3. The `Sorting` used in `Step-4` can be done in `O(n)` instead of `O(nlogn)` as only one digit needs to be repositioned else all are already sorted.
4. This `Optimization` would improve our solution from `O(n+n+nlogn)` to `O(n+logn+n)` giving overall time complexity of `O(n)`, where
`n = number-of-digits-of-a-given-number`.

SOLUTIONS:

```#include <stdio.h>

//function for swapping two values
void swap(int arr[], int i, int j)
{
int temp = arr[i];
arr[i] = arr[j];
arr[j] = temp;
}

//sorting a subarray in increasing order
void sortSubarray(int number[], int i, int j)
{
while (i < j)
{
swap(number, i, j);
i += 1;
j -= 1;
}
}

void findNextGreaterNumber(int arr[], int n)
{
int lastDigitSeen = arr[n - 1], i, j;
for (i = n - 2; i >= 0; i--)
{
if (lastDigitSeen > arr[i])
break;
lastDigitSeen = arr[i];
}
if (i >= 0)
{
for (j = n - 1; j > i; j--)
{
if (arr[j] > arr[i])
break;
}

swap(arr, i, j);
sortSubarray(arr, i + 1, n - 1);
}
}
int main()
{
int t;
scanf("%d", &t);
while (t--)
{
long long n;
scanf("%lld", &n);
int count = 0;
long long num = n;

//count the number of digits in n
while (num > 0)
{
num /= 10;
count++;
}

//storing all digits of n into an array
int arr[count];
int i = count - 1;
while (n > 0)
{
arr[i--] = n % 10;
n /= 10;
}

findNextGreaterNumber(arr, count);

//printing all digits of the resultant number
for (int j = 0; j < count; j++)
printf("%d", arr[j]);
printf("\n");
}
return 0;
}
```
```
#include <bits/stdc++.h>
using namespace std;

//function for swapping two values
void swap(int arr[], int i, int j)
{
int temp = arr[i];
arr[i] = arr[j];
arr[j] = temp;
}

//sorting a subarray in increasing order
void sortSubarray(int number[], int i, int j)
{
while (i < j)
{
swap(number, i, j);
i += 1;
j -= 1;
}
}

void findNextGreaterNumber(int arr[], int n)
{
int lastDigitSeen = arr[n - 1], i, j;
for (i = n - 2; i >= 0; i--)
{
if (lastDigitSeen > arr[i])
break;
lastDigitSeen = arr[i];
}
if (i >= 0)
{
for (j = n - 1; j > i; j--)
{
if (arr[j] > arr[i])
break;
}

swap(arr, i, j);
sortSubarray(arr, i + 1, n - 1);
}
}
int main()
{
int t;
cin >> t;
while (t--)
{
long long n;
cin >> n;
int count = 0;
long long num = n;

//count the number of digits in n
while (num > 0)
{
num /= 10;
count++;
}

//storing all digits of n into an array
int arr[count];
int i = count - 1;
while (n > 0)
{
arr[i--] = n % 10;
n /= 10;
}

findNextGreaterNumber(arr, count);

//printing all digits of the resultant number
for (int j = 0; j < count; j++)
cout << arr[j];
cout << "\n";
}
return 0;
}
```
```
#include <bits/stdc++.h>
using namespace std;

string NextGreater(string str){
int n = str.size();
for(int i=n-2; i>=0; i--){
if(str[i] < str[i+1]){
sort(str.begin() + i + 1, str.end());
auto it = upper_bound(str.begin() + i + 1, str.end(), str[i]);
int index = it - str.begin();
swap(str[index], str[i]);
break;
}
}
return str;
}
int main()
{
int t; cin>>t;
while(t--){
string str; cin>>str;
cout<<"\n"<<NextGreater(str);
}
return 0;
}
```
```
import java.util.*;
import java.io.*;

public class Main {
static void swap(long []arr, int i, int j)
{
long temp = arr[i];
arr[i] = arr[j];
arr[j] = temp;
}

//sorting a subarray in increasing order
static void sortSubarray(long []number, int i, int j)
{
while (i < j)
{
swap(number, i, j);
i += 1;
j -= 1;
}
}

static void findNextGreaterNumber(long []arr, int n)
{
long lastDigitSeen = arr[n - 1];
int i, j;
for (i = n - 2; i >= 0; i--)
{
if (lastDigitSeen > arr[i])
break;
lastDigitSeen = arr[i];
}
if (i >= 0)
{
for (j = n - 1; j > i; j--)
{
if (arr[j] > arr[i])
break;
}

swap(arr, i, j);
sortSubarray(arr, i + 1, n - 1);
}
}
public static void main(String args[]) throws IOException {

Scanner sc = new Scanner(System.in);
int t = sc.nextInt();
while (t != 0)
{
long n = sc.nextLong();
int count = 0;
long num = n;

//count the number of digits in n
while (num > 0)
{
num /= 10;
count++;
}

//storing all digits of n into an array
long arr[] = new long[count];
int i = count - 1;
while (n > 0)
{
arr[i--] = n % 10;
n /= 10;
}

findNextGreaterNumber(arr, count);

//printing all digits of the resultant number
for (int j = 0; j < count; j++)
System.out.print(arr[j]);
System.out.println();
t--;
}

}
}
```

[forminator_quiz id="1657"]
where `N = Number of digits in the number`.
Space Complexity: O(1)

This article tried to discuss Searching. Hope this blog helps you understand and solve the problem. To practice more problems on Searching you can check out MYCODE | Competitive Programming.