Positions of Large Groups

Positions of Large Groups - Problem

String Easy

In a string s of lowercase letters, these letters form consecutive groups of the same character.

For example, a string like s = "abbxxxxzyy" has the groups "a", "bb", "xxxx", "z", and "yy".

A group is identified by an interval [start, end], where start and end denote the start and end indices (inclusive) of the group. In the above example, "xxxx" has the interval [3,6].

A group is considered large if it has 3 or more characters.

Return the intervals of every large group sorted in increasing order by start index.

Input & Output

Example 1 — Basic Case

$ Input: s = "abbxxxxzyy"

› Output: [[3,6]]

💡 Note: The groups are "a" (length 1), "bb" (length 2), "xxxx" (length 4), "z" (length 1), "yy" (length 2). Only "xxxx" has length ≥ 3, spanning indices [3,6].

Example 2 — Multiple Large Groups

$ Input: s = "abcdddeeeeaabbbcd"

› Output: [[3,5],[6,9],[12,14]]

💡 Note: Groups: "a"(1), "b"(1), "c"(1), "ddd"(3), "eeee"(4), "aa"(2), "bbb"(3), "c"(1), "d"(1). Large groups are "ddd" [3,5], "eeee" [6,9], "bbb" [12,14].

Example 3 — No Large Groups

$ Input: s = "abc"

› Output: []

💡 Note: All groups have length 1, which is less than 3, so no large groups exist.

Constraints

1 ≤ s.length ≤ 1000
s consists of lowercase English letters only

Visualization

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Asked in

G Google 12 f Facebook 8

The key insight is to scan the string once while tracking consecutive groups. Best approach uses single pass scanning to identify when characters change and check if the previous group was large (≥3 characters). Time: O(n), Space: O(1)

Common Approaches

✓ Brute Force - Check All Substrings

⏱️ Time: O(n²) Space: O(1)

For each position, try to extend a group as far as possible by checking all characters. This involves nested loops to examine every potential group starting position.

Single Pass Scan

⏱️ Time: O(n) Space: O(1)

Use a single pass through the string, keeping track of the current group's start position and length. When we encounter a different character, check if the previous group was large.

Brute Force - Check All Substrings — Algorithm Steps

For each starting position i
Try to extend group as long as characters match
If group length ≥ 3, add [start, end] to result

Visualization

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Step-by-Step Walkthrough

Start at position 0

Try to extend 'a' group

Move to next different char

Check 'bb' group (length 2, not large)

Find large group

'xxxx' has length 4 ≥ 3, so add [3,6]

Code -

solution.c — C

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

void solution(char* s, int result[][2], int* resultSize) {
    *resultSize = 0;
    int n = strlen(s);
    
    for (int i = 0; i < n; i++) {
        int start = i;
        // Extend group as far as possible
        while (i < n - 1 && s[i] == s[i + 1]) {
            i++;
        }
        
        // Check if group is large (3+ characters)
        if (i - start + 1 >= 3) {
            result[*resultSize][0] = start;
            result[*resultSize][1] = i;
            (*resultSize)++;
        }
    }
}

int main() {
    char s[1000];
    fgets(s, sizeof(s), stdin);
    s[strcspn(s, "\n")] = 0;  // Remove newline
    
    int result[100][2];
    int resultSize;
    
    solution(s, result, &resultSize);
    
    printf("[");
    for (int i = 0; i < resultSize; i++) {
        if (i > 0) printf(",");
        printf("[%d,%d]", result[i][0], result[i][1]);
    }
    printf("]\n");
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n²)

Nested loops: outer loop n positions, inner loop up to n characters

✓ Linear Growth

Space Complexity

O(1)

Only using variables for tracking, result array doesn't count toward space complexity

✓ Linear Space

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Input & Output

Constraints

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Related Problems

Common Approaches

Brute Force - Check All Substrings — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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