Skip to content
Related Articles

Related Articles

Improve Article

Lexicographically largest string using at most K swaps at same parity indices

  • Last Updated : 05 Jul, 2021

Given string S and a positive integer K, the task is to find lexicographically the largest possible string using at most K swaps with the condition that the indices that are swapped must be either both odd or both even.

Examples: 

Input: S = “ancqz”, K = 2
Output:  “zqcna
Explanation: In one swap, we can swap characters ‘n’ and ‘q’ as they both are at even indices (2 and 4 assuming 1-based indexing). The string becomes “aqcnz”. In the second swap we can swap characters ‘a’ and ‘z’ as both have odd indices. The final string “zqcna” is the largest lexicographically possible using 2 swap operations.

Note: We cannot swap for instance ‘a’ and ‘n’ or ‘n’ and ‘z’ as one of them would be at an odd index while the other at even index.  

Input: S = “geeksforgeeks”, K = 3
Output:  “sreksfoegeekg



Naive approach: The naive approach is trivial. Use the greedy algorithm to make the current index maximum starting from left by picking the maximum possible character that is to the right of the current index and is also with the same parity of index i.e. (odd if the current index is odd and even if the current index is even). Repeat the same procedure atmost K times. The time complexity of the approach will be O(N2).

Efficient Approach: The above approach can be improved using a priority queue. Follow the steps below to solve the problem:

  • Create two priority queue one for odd index characters and another for even index characters.
  • Iterate over characters in the string, if the even index character comes then search the index which is greater than the current index and character bigger than the current character in the priority queue which is holding even characters. If there is any, swap the two characters push the current character and the index that we found in the priority queue.
  • The same procedure is to be followed when the odd character comes.
  • If K becomes 0, terminate the loop.
  • The resultant string will be the answer.

Below is the implementation of the above approach:

C++




// C++ program of the above approach
#include <bits/stdc++.h>
using namespace std;
  
// Function which returns
// the largest possible string
string lexicographicallyLargest(string S, int K)
{
    // Finding length of the string
    int n = S.length();
  
    // Creating two priority queues of pairs
    // for odd and even indices separately
    priority_queue<pair<char, int> > pqOdd, pqEven;
  
    // Storing all possible even
    // indexed values as pairs
    for (int i = 2; i < n; i = i + 2) {
        // Stores pair as {character, index}
        pqEven.push(make_pair(S[i], i));
    }
  
    // Storing all possible odd indexed
    // values as pairs
    for (int i = 3; i < n; i = i + 2) {
        // Stores pair as {character, index}
        pqOdd.push(make_pair(S[i], i));
    }
  
    for (int i = 0; i < n; i++) {
        // For even indices
        if (i % 2 == 0) {
  
            // Removing pairs which
            // cannot be used further
            while (!pqEven.empty()
                   and pqEven.top().second <= i)
                pqEven.pop();
  
            // If a pair is found whose index comes after
            // the current index and its character is
            // greater than the current character
            if (!pqEven.empty()
                and pqEven.top().first > S[i]) {
  
                // Swap the current index with index of
                // maximum found character next to it
                swap(S[i], S[pqEven.top().second]);
  
                int idx = pqEven.top().second;
                pqEven.pop();
                // Push the updated character at idx index
                pqEven.push({ S[idx], idx });
                K--;
            }
        }
        // For odd indices
        else {
            // Removing pairs which cannot
            // be used further
  
            while (!pqOdd.empty()
                   and pqOdd.top().second <= i)
                pqOdd.pop();
  
            // If a pair is found whose index comes after
            // the current index and its character is
            // greater than the current character
  
            if (!pqOdd.empty()
                and pqOdd.top().first > S[i]) {
  
                // Swap the current index with index of
                // maximum found character next to it
                swap(S[i], S[pqOdd.top().second]);
                int idx = pqOdd.top().second;
                pqOdd.pop();
  
                // Push the updated character at idx index
                pqOdd.push({ S[idx], idx });
                K--;
            }
        }
        // Breaking out of the loop if K=0
        if (K == 0)
            break;
    }
  
    return S;
}
  
// Driver Code
int main()
{
    // Input
    string S = "geeksforgeeks";
    int K = 2;
  
    // Function Call
    cout << lexicographicallyLargest(S, K);
    return 0;
}
Output:
sreksfoegeekg

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

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.




My Personal Notes arrow_drop_up
Recommended Articles
Page :