Next Greater Element III in C++

C++Server Side ProgrammingProgramming

Suppose we have a positive 32-bit integer n, we need to find the smallest 32-bit integer which has exactly the same digits existing in the integer n and is greater in value than n. If we have no such positive 32-bit integer number, then return -1.

So if the number is 213, then the result will be 231.

To solve this, we will follow these steps −

  • s := n as string, sz := size of s, ok := false
  • for i in range sz – 2 to 0
    • if s[i] < s[i + 1], then ok := true and break the loop
  • if of is false, then return – 1
  • smallest := i, curr := i + 1
  • for j in range i + 1 to sz – 1
    • id s[j] > s[smallest] and s[j] <= s[curr], then curr := j
  • exchange s[smallest] with s[curr]
  • aux := substring of s from index 0 to smallest
  • reverse aux
  • ret := substring of s from index 0 to smallest + aux
  • return -1 if ret is > 32-bit +ve integer range, otherwise ret

Let us see the following implementation to get better understanding −

Example

 Live Demo

#include <bits/stdc++.h>
using namespace std;
class Solution {
   public:
   int nextGreaterElement(int n) {
      string s = to_string(n);
      int sz = s.size();
      int i;
      bool ok = false;
      for(i = sz - 2; i >= 0; i--){
         if(s[i] < s[i + 1]) {
            ok = true;
            break;
         }
      }
      if(!ok) return -1;
      int smallest = i;
      int curr = i + 1;
      for(int j = i + 1; j < sz; j++){
         if(s[j] > s[smallest] && s[j] <= s[curr]){
            curr = j;
         }
      }
      swap(s[smallest], s[curr]);
      string aux = s.substr(smallest + 1);
      reverse(aux.begin(), aux.end());
      string ret = s.substr(0, smallest + 1) + aux;
      return stol(ret) > INT_MAX ? -1 : stol(ret);
   }
};
main(){
   Solution ob;
   cout << (ob.nextGreaterElement(213));
}

Input

213

Output

231
raja
Updated on 04-May-2020 13:09:59

Advertisements