Java Program to Swap characters in a String

• Difficulty Level : Medium
• Last Updated : 25 Jan, 2022

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:

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;
Explanation:
after 1st swap: BACDE
after 3rd swap: BCDAE
after 4th swap: BCDEA
after 5th swap: ACDEB
after 7th swap: CDAEB
after 8th swap: CDEAB
after 9th swap: CDEBA

Naive Approach

• 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.

Efficient Approach:

• 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 IJKEFGH
after 2 full iteration: CDAB FGHIJKE
after 3 full iteration: BCDA JKEFGHI
after 4 full iteration: ABCD GHIJKEF
after 5 full iteration: DABC KEFGHIJ
after 6 full iteration: CDAB HIJKEFG
after 7 full iteration: BCDA EFGHIJK
after 8 full iteration: ABCD IJKEFGH

Below is the implementation of the approach:

Java

 // Java Program to find new after swapping // characters at position i and i + c // b times, each time advancing one // position ahead    class GFG {     // Method to find the required string        String swapChars(String s, int c, int b)     {         // Get string length         int n = s.length();            // 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.substring(0, c),                                ((n % c) * f) % c);            // Rotate remaining character by         // (n * f) places         String p2 = rotateLeft(s.substring(c),                                ((c * f) % (n - c)));            // Concatenate the two parts and convert the         // resultant string formed after f full         // iterations to a character array         // (for final swaps)         char a[] = (p1 + p2).toCharArray();            // 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 new String(a);     }        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.substring(p) + s.substring(0, p);     }        // Driver code     public static void main(String args[])     {         // Given values         String s1 = "ABCDEFGHIJK";         int b = 1000;         int c = 3;            // get final string         String s2 = new GFG().swapChars(s1, c, b);            // print final string         System.out.println(s2);     } }

Output:

Time Complexity: O(n)
Space Complexity: O(n)

Please refer complete article on Swap characters in a String for more details!

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