New Line – Repeated Characters

Split a string into substrings at positions where characters occur repeatedly

Understand the Problem

Problem Statement

The program must accept a string S as the input. The program must split the string S wherever a character occurs repeatedly(i.e., two or more occurrences). Then the program must print the substrings as the output.

Constraints

2 <= Length of S <= 100
String S contains only alphabetic characters (case-sensitive)
Each character can appear multiple times in the string
Output substrings must be printed on separate lines

Examples

Example 1

Input

skillrack

Output

skil
lrac
k

Explanation

The string 'skillrack' is split at positions where characters repeat. The character 'l' appears at positions 3 and 4 (0-indexed), and 'k' appears at positions 0 and 8. The substrings are: 'skil' (before first 'l'), 'lrac' (between 'l' and 'k'), and 'k' (after the last 'k').

Example 2

Input

ENvironment

Output

ENvironme
nt

Explanation

The string 'ENvironment' has the character 'n' appearing at positions 1 and 8 (0-indexed). The string is split into 'ENvironme' (before the repeated 'n') and 'nt' (after the repeated 'n').

Example 3

Input

abcdabcdefge

Output

abcd
a
b
c
defg
e

Explanation

The string 'abcdabcdefge' has multiple repeated characters: 'a' at positions 0 and 4, 'b' at positions 1 and 5, 'c' at positions 2 and 6, and 'e' at positions 7 and 11. The string is split at each repeated character position, creating substrings: 'abcd', 'a', 'b', 'c', 'defg', and 'e'.

Solution

#include <stdio.h>
#include <string.h>

int main() {
    char s[101];
    fgets(s, sizeof(s), stdin);
    
    // Remove newline character if present
    int len = strlen(s);
    if (s[len-1] == '\n') {
        s[len-1] = '\0';
        len--;
    }
    
    int splitPoints[100];
    int splitCount = 0;
    int charCount[256] = {0}; // ASCII character frequency
    
    // Find split points where characters repeat
    for (int i = 0; i < len; i++) {
        charCount[(unsigned char)s[i]]++;
        if (charCount[(unsigned char)s[i]] == 2) {
            splitPoints[splitCount++] = i;
        }
    }
    
    // Print substrings
    int start = 0;
    for (int i = 0; i < splitCount; i++) {
        for (int j = start; j < splitPoints[i]; j++) {
            printf("%c", s[j]);
        }
        printf("\n");
        start = splitPoints[i];
    }
    
    // Print last substring
    for (int j = start; j < len; j++) {
        printf("%c", s[j]);
    }
    printf("\n");
    
    return 0;
}

Time:O(n) where n is the length of the string

Space:O(1) for the fixed-size arrays (256 for charCount, 100 for splitPoints)

Approach:

Step-by-step explanation:

Read the input string using fgets() and remove trailing newline
Create an array splitPoints to store positions where splits should occur
Use a frequency array charCount to track character occurrences
Iterate through the string, incrementing character counts
When a character count reaches 2, record its position as a split point
Print substrings by iterating between consecutive split points
Handle the final substring from the last split point to the end

The solution efficiently tracks character frequencies using an array indexed by ASCII values, ensuring O(1) lookup time for character counting.

Visual Explanation

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