Given a String S of length N, two integers B and C, the task is to traverse characters starting from the beginning, swapping a character with the character after C places from it, i.e. swap characters at position i and (i + C)%N. Repeat this process B times, advancing one position at a time. Your task is to find the final String after B swaps.
Examples:Â
Input : S = "ABCDEFGH", B = 4, C = 3;
Output: DEFGBCAH
Explanation:
after 1st swap: DBCAEFGH
after 2nd swap: DECABFGH
after 3rd swap: DEFABCGH
after 4th swap: DEFGBCAH
Input : S = "ABCDE", B = 10, C = 6;
Output : ADEBC
Explanation:
after 1st swap: BACDE
after 2nd swap: BCADE
after 3rd swap: BCDAE
after 4th swap: BCDEA
after 5th swap: ACDEB
after 6th swap: CADEB
after 7th swap: CDAEB
after 8th swap: CDEAB
after 9th swap: CDEBA
after 10th swap: ADEBC
Naive Approach:Â
- For large values of B, the naive approach of looping B times, each time swapping ith character with (i + C)%N-th character will result in high CPU time.
- The trick to solving this problem is to observe the resultant string after every N iterations, where N is the length of the string S.
- Again, if C is greater than or equal to the N, it is effectively equal to the remainder of C divided by N.
- Hereon, let’s consider C to be less than N.
Below is the implementation of the approach:
C++
// C++ Program to Swap characters in a String #include <iostream> #include <string>   using namespace std;   string swapCharacters(string s, int B, int C) {     int N = s.size();     // If c is greater than n     C = C % N;     // loop to swap ith element with (i + C) % n th element     for (int i = 0; i < B; i++) {         swap(s[i % N], s[(i + C) % N]);     }     return s; }   int main() {     string s = "ABCDEFGH";     int B = 4;     int C = 3;     s = swapCharacters(s, B, C);     cout << s << endl;     return 0; }   // This code is contributed by Susobhan Akhuli |
Output
DEFGBCAH
Time Complexity: O(B), to iterate B times.
Space Complexity: O(1)
Efficient Approach:Â
- If we observe the string that is formed after every N successive iterations and swaps (let’s call it one full iteration), we can start to get a pattern.
- We can find that the string is divided into two parts: the first part of length C comprising of the first C characters of S, and the second part comprising of the rest of the characters.
- The two parts are rotated by some places. The first part is rotated right by (N % C) places every full iteration.
- The second part is rotated left by C places every full iteration.
- We can calculate the number of full iterations f by dividing B by N.
- So, the first part will be rotated left by ( N % C ) * f . This value can go beyond C and so, it is effectively ( ( N % C ) * f ) % C, i.e. the first part will be rotated by ( ( N % C ) * f ) % C places left.
- The second part will be rotated left by C * f places. Since, this value can go beyond the length of the second part which is ( N – C ), it is effectively ( ( C * f ) % ( N – C ) ), i.e. the second part will be rotated by ( ( C * f ) % ( N – C ) ) places left.
- After f full iterations, there may still be some iterations remaining to complete B iterations. This value is B % N which is less than N. We can follow the naive approach on these remaining iterations after f full iterations to get the resultant string.
Example:
s = ABCDEFGHIJK; c = 4;
parts: ABCD EFGHIJK
after 1 full iteration: DABC IJKEFGHÂ
after 2 full iteration: CDAB FGHIJKEÂ
after 3 full iteration: BCDA JKEFGHIÂ
after 4 full iteration: ABCD GHIJKEFÂ
after 5 full iteration: DABC KEFGHIJÂ
after 6 full iteration: CDAB HIJKEFGÂ
after 7 full iteration: BCDA EFGHIJKÂ
after 8 full iteration: ABCD IJKEFGHÂ
Â
Below is the implementation of the approach:
C++
// C++ program to find new after swapping // characters at position i and i + c // b times, each time advancing one // position ahead #include <bits/stdc++.h> using namespace std;   string rotateLeft(string s, int p) {           // Rotating a string p times left is     // effectively cutting the first p     // characters and placing them at the end     return s.substr(p) + s.substr(0, p); }   // Method to find the required string string swapChars(string s, int c, int b) {           // Get string length     int n = s.size();           // If c is larger or equal to the length of     // the string is effectively the remainder of     // c divided by the length of the string     c = c % n;           if (c == 0)     {                   // No change will happen         return s;     }     int f = b / n;     int r = b % n;           // Rotate first c characters by (n % c)     // places f times     string p1 = rotateLeft(s.substr(0, c),                   ((n % c) * f) % c);                         // Rotate remaining character by     // (n * f) places     string p2 = rotateLeft(s.substr(c),                   ((c * f) % (n - c)));                         // Concatenate the two parts and convert the     // resultant string formed after f full     // iterations to a string array     // (for final swaps)     string a = p1 + p2;           // Remaining swaps     for(int i = 0; i < r; i++)     {                   // Swap ith character with         // (i + c)th character         char temp = a[i];         a[i] = a[(i + c) % n];         a[(i + c) % n] = temp;     }           // Return final string     return a; }   // Driver code int main() {           // Given values     string s1 = "ABCDEFGHIJK";     int b = 1000;     int c = 3;           // Get final string print final string     cout << swapChars(s1, c, b) << endl; }   // This code is contributed by rag2127 |
Output:Â
CADEFGHIJKB
Â
Time Complexity: O(n)Â
Space Complexity: O(n)
Â
Please refer complete article on Swap characters in a String for more details!
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
… [Trackback]
[…] Find More to that Topic: geeksforgeeks.org/c-program-to-swap-characters-in-a-string/ […]
… [Trackback]
[…] Find More to that Topic: geeksforgeeks.org/c-program-to-swap-characters-in-a-string/ […]