Find Maximum Number of Non Intersecting Substrings

Find Maximum Number of Non Intersecting Substrings - Problem

You are given a string word. Return the maximum number of non-intersecting substrings of word that are at least four characters long and start and end with the same letter.

A substring is non-intersecting if it does not overlap with any other selected substring. The goal is to maximize the count of such valid substrings.

Input & Output

Example 1 — Basic Case

$ Input: word = "ababa"

› Output: 1

💡 Note: Valid substrings: "abab" (0-3), "baba" (1-4), "ababa" (0-4). Only one can be selected since they all overlap. Maximum count is 1.

Example 2 — Multiple Non-overlapping

$ Input: word = "abcabc"

› Output: 1

💡 Note: Valid substrings: "abca" (0-3), "bcab" (1-4), "cabc" (2-5). They all overlap, so maximum count is 1.

Example 3 — No Valid Substrings

$ Input: word = "abcd"

› Output: 0

💡 Note: No substring of length ≥ 4 starts and ends with the same character. Return 0.

Constraints

1 ≤ word.length ≤ 1000
word consists of lowercase English letters

Visualization

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Asked in

G Google 25 a Amazon 20 M Microsoft 15

The key insight is to treat this as an interval scheduling problem. Find all valid substrings (length ≥ 4, same start/end char), then use greedy selection by sorting intervals by end position. Best approach is Greedy with Time: O(n² log n), Space: O(n²)

Common Approaches

✓ Heap

⏱️ Time: N/A Space: N/A

Brute Force - Try All Combinations

⏱️ Time: O(2^n) Space: O(n²)

Find all valid substrings (length ≥ 4, same start/end char), then use backtracking to try every possible combination of non-overlapping substrings to find the maximum count.

Dynamic Programming

⏱️ Time: O(n²) Space: O(n)

For each position, calculate the maximum number of non-overlapping valid substrings that can end at or before that position, considering all possible valid substrings.

Greedy - Select Earliest Ending

⏱️ Time: O(n² log n) Space: O(n²)

Find all valid substrings, sort them by ending position, then greedily select substrings that don't overlap with previously selected ones. This maximizes the count.

Algorithm Steps — Algorithm Steps

Code -

solution.c — C

Time & Space Complexity

Time Complexity

⏱️

⚠ Quadratic Growth

Space Complexity

⚠ Quadratic Space

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Smart Actions

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