Greater And Least






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"


  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.


  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 ithindex 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.


  1. A few optimizations can be made in this approach and significantly improve the Time Complexity of the approach according to the data structures and algorithms in c++.

  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.


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