LeetCode #830 — EASY

Positions of Large Groups

Build confidence with an intuition-first walkthrough focused on core interview patterns fundamentals.

The Problem

Problem Statement

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.

Example 1:

Input: s = "abbxxxxzzy"
Output: [[3,6]]
Explanation: "xxxx" is the only large group with start index 3 and end index 6.

Example 2:

Input: s = "abc"
Output: []
Explanation: We have groups "a", "b", and "c", none of which are large groups.

Example 3:

Input: s = "abcdddeeeeaabbbcd"
Output: [[3,5],[6,9],[12,14]]
Explanation: The large groups are "ddd", "eeee", and "bbb".

Constraints:

1 <= s.length <= 1000
s contains lowercase English letters only.

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: General problem-solving

Example 1

"abbxxxxzzy"

Example 2

"abc"

Example 3

"abcdddeeeeaabbbcd"

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #830: Positions of Large Groups
class Solution {
    public List<List<Integer>> largeGroupPositions(String s) {
        int n = s.length();
        int i = 0;
        List<List<Integer>> ans = new ArrayList<>();
        while (i < n) {
            int j = i;
            while (j < n && s.charAt(j) == s.charAt(i)) {
                ++j;
            }
            if (j - i >= 3) {
                ans.add(List.of(i, j - 1));
            }
            i = j;
        }
        return ans;
    }
}

// Accepted solution for LeetCode #830: Positions of Large Groups
func largeGroupPositions(s string) [][]int {
	i, n := 0, len(s)
	ans := [][]int{}
	for i < n {
		j := i
		for j < n && s[j] == s[i] {
			j++
		}
		if j-i >= 3 {
			ans = append(ans, []int{i, j - 1})
		}
		i = j
	}
	return ans
}

# Accepted solution for LeetCode #830: Positions of Large Groups
class Solution:
    def largeGroupPositions(self, s: str) -> List[List[int]]:
        i, n = 0, len(s)
        ans = []
        while i < n:
            j = i
            while j < n and s[j] == s[i]:
                j += 1
            if j - i >= 3:
                ans.append([i, j - 1])
            i = j
        return ans

// Accepted solution for LeetCode #830: Positions of Large Groups
struct Solution;

struct Group {
    c: char,
    start: usize,
    end: usize,
}

impl Solution {
    fn large_group_positions(s: String) -> Vec<Vec<i32>> {
        let mut prev: Option<Group> = None;
        let mut groups: Vec<Group> = vec![];
        for (i, c) in s.chars().enumerate() {
            if let Some(prev_group) = prev {
                if prev_group.c == c {
                    prev = Some(Group {
                        c,
                        start: prev_group.start,
                        end: i,
                    });
                } else {
                    groups.push(prev_group);
                    prev = Some(Group {
                        c,
                        start: i,
                        end: i,
                    });
                }
            } else {
                prev = Some(Group {
                    c,
                    start: i,
                    end: i,
                });
            }
        }
        if let Some(prev_group) = prev {
            groups.push(prev_group);
        }

        groups
            .iter()
            .filter_map(|g| {
                let start = g.start;
                let end = g.end;
                if end - start > 1 {
                    Some(vec![start as i32, end as i32])
                } else {
                    None
                }
            })
            .collect()
    }
}

#[test]
fn test() {
    let s = "abbxxxxzzy".to_string();
    let res: Vec<Vec<i32>> = vec_vec_i32![[3, 6]];
    assert_eq!(Solution::large_group_positions(s), res);
    let s = "abc".to_string();
    let res: Vec<Vec<i32>> = vec![];
    assert_eq!(Solution::large_group_positions(s), res);
    let s = "abcdddeeeeaabbbcd".to_string();
    let res: Vec<Vec<i32>> = vec_vec_i32![[3, 5], [6, 9], [12, 14]];
    assert_eq!(Solution::large_group_positions(s), res);
}

// Accepted solution for LeetCode #830: Positions of Large Groups
function largeGroupPositions(s: string): number[][] {
    const n = s.length;
    const ans: number[][] = [];

    for (let i = 0; i < n; ) {
        let j = i;
        while (j < n && s[j] === s[i]) {
            ++j;
        }
        if (j - i >= 3) {
            ans.push([i, j - 1]);
        }
        i = j;
    }

    return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(1)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.