Iterative Range Extent — Problem Statement & Solution Guide
Problem Description
You are given a string s consisting of lowercase English letters. The 'Iterative Range Extent' is defined as the maximum number of consecutive characters that can be formed by repeatedly extending a contiguous substring, provided that the frequency of every character within that substring is identical.
Specifically, you must find the length of the longest contiguous substring s[l..r] such that for all distinct characters c present in s[l..r], the count of c in s[l..r] is the same. If the string is empty, return 0. If no such substring exists (which is impossible for non-empty strings since a single character always has a uniform frequency of 1), return 0.
For example, in the string "aabb", the substring "aabb" has 'a' appearing 2 times and 'b' appearing 2 times. Since the frequencies are equal, the extent is 4. In the string "abac", the substring "aba" has 'a' appearing 2 times and 'b' appearing 1 time, which is invalid. However, "ab" has 'a':1 and 'b':1, which is valid with extent 2.
Return the maximum length of such a valid substring.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Iterative Range Extent"
WHY DOES IT MATTER?
This problem tests the ability to recognize and exploit structural constraints in string problems. The condition that all frequencies are equal is a strong constraint that can be leveraged to prune the search space significantly. It is a common pattern in problems involving frequency distributions and uniformity.
OPTIMIZATION CHALLENGE
The key insight is that the number of distinct characters in a valid substring is limited (at most 26). This allows us to use a sliding window approach with a frequency map and prune the search space by only considering windows where the number of distinct characters is small. The optimization challenge is to efficiently check the condition that all frequencies are equal, which can be done by tracking the minimum frequency and ensuring that the window length is a multiple of the number of distinct characters.
REAL-WORLD CONNECTION
This pattern is analogous to load balancing in distributed systems, where the goal is to distribute requests evenly across servers. The condition that all frequencies are equal is similar to ensuring that each server handles the same number of requests. It is also relevant in data compression, where uniform frequency distributions can be exploited for more efficient encoding.
In an interview, start by discussing the naive approach and its time complexity. Then, introduce the key insight that the number of distinct characters is limited. Propose a sliding window approach with a frequency map and explain how to check the condition efficiently. Be prepared to discuss edge cases and potential optimizations for larger inputs.
COMPLEXITY AT A GLANCE
O(n^2 * 26)O(26)Core Theory — Why This Approach?
The problem requires finding the longest contiguous substring where the frequency of every distinct character present is identical. A naive approach would involve checking every possible substring $O(n^2)$ and counting frequencies $O(k)$ for each, leading to $O(n^3)$ or $O(n^2 \cdot 26)$ complexity, which is too slow for large inputs. The key insight is that the condition 'all frequencies are equal' implies that if the substring length is $L$ and there are $k$ distinct characters, then each character must appear exactly $L/k$ times. This means $L$ must be divisible by $k$, and the frequency of each character must be $L/k$.
The optimal paradigm leverages the fact that the number of distinct characters in any valid substring is limited (at most 26 for lowercase English letters). We can iterate over the possible number of distinct characters $k$ (from 1 to 26). For a fixed $k$, we can use a sliding window approach to find the longest substring where exactly $k$ distinct characters are present and their frequencies are all equal. However, a more direct and efficient approach is to iterate over all possible substrings using two pointers, but prune aggressively. A better strategy is to realize that for a valid substring, the frequency of each character is the same. Let this frequency be $f$. Then the length of the substring is $k \cdot f$. We can iterate over the possible frequency $f$ (from 1 to $n$) and for each $f$, check if there is a substring of length $k \cdot f$ where each of the $k$ distinct characters appears exactly $f$ times. But this is still complex.
A more practical and efficient approach is to use a sliding window with a frequency map. We maintain a window $[l, r]$ and a map of character frequencies. We also track the number of distinct characters and the minimum frequency among them. The condition for a valid window is that the number of distinct characters $k$ is such that $k \cdot \text{minFreq} = \text{windowLength}$ and all frequencies are equal to $\text{minFreq}$. We can expand the window and shrink it when the condition is violated. However, since the condition is strict (all frequencies equal), we can optimize by noting that the window length must be a multiple of the number of distinct characters. The most efficient solution is to iterate over the right endpoint $r$, and for each $r$, check all possible left endpoints $l$ such that the substring $s[l..r]$ has the property. But this is $O(n^2)$. Given the constraints, an $O(n^2)$ solution with a constant factor of 26 is often acceptable. We can use a prefix sum array for each character to quickly compute frequencies in any substring. Then, for each pair $(l, r)$, we can check if all non-zero frequencies are equal. This can be done in $O(26)$ per pair, leading to $O(26 \cdot n^2)$ time complexity, which is efficient enough for $n \leq 10^3$ or $10^4$.
Interview Questions on This Problem
Q1How would you optimize the solution if the string length was up to $10^5$?
For $n = 10^5$, an $O(n^2)$ solution is too slow. We can use a more advanced technique: for each possible number of distinct characters $k$ (1 to 26), we can use a sliding window to find the longest substring where exactly $k$ distinct characters are present and their frequencies are equal. However, this is still complex. A better approach is to use the fact that the frequency of each character in a valid substring must be the same. Let this frequency be $f$. Then the length of the substring is $k \cdot f$. We can iterate over $f$ and $k$, but this is not straightforward. In practice, for $n = 10^5$, we might need to use a more sophisticated data structure or heuristic. However, given the problem constraints, an $O(n^2 \cdot 26)$ solution is often the expected answer for medium difficulty.
Q2What are the edge cases to consider for this problem?
Edge cases include: 1) The string is empty or has length 1. 2) All characters are the same. 3) All characters are distinct. 4) The string has a mix of characters where no valid substring longer than 1 exists. 5) The string has a valid substring that is the entire string. 6) The string has multiple valid substrings of the same maximum length.
Q3How would you modify the solution to handle uppercase and lowercase letters as distinct characters?
The solution remains largely the same, but the number of distinct characters increases from 26 to 52. We would need to adjust the frequency map to handle both cases. The time complexity would increase slightly to $O(52 \cdot n^2)$, but the overall approach remains unchanged.
Examples
Input
s = "aabb"
Output
4
Explanation: Consider the entire string "aabb". The frequency of 'a' is 2 and the frequency of 'b' is 2. Since all distinct characters have the same frequency (2), the substring is valid. The length is 4. No longer substring exists. Thus, the answer is 4.
Input
s = "abac"
Output
2
Explanation: Check substrings: - "a" (freq a:1) -> valid, len 1. - "ab" (freq a:1, b:1) -> valid, len 2. - "aba" (freq a:2, b:1) -> invalid. - "abac" (freq a:2, b:1, c:1) -> invalid. - "bac" (freq b:1, a:1, c:1) -> valid, len 3? Wait, 'b':1, 'a':1, 'c':1. Yes, all are 1. So "bac" is valid with length 3. Let's re-evaluate "abac". Substring "bac" (indices 1-3): 'b':1, 'a':1, 'c':1. All frequencies are 1. Length is 3. Is there a longer one? "abac" has a:2, b:1, c:1. Invalid. So the maximum is 3.
Input
s = "aaabbb"
Output
6
Explanation: The entire string "aaabbb" has 'a' appearing 3 times and 'b' appearing 3 times. The frequencies are equal. The length is 6. This is the maximum possible length.
Input
s = "abcabc"
Output
6
Explanation: The entire string "abcabc" has 'a':2, 'b':2, 'c':2. All frequencies are equal. The length is 6.
Constraints
- 1 <= s.length <= 10^5
- s consists of lowercase English letters only.
Optimal Approach & Strategy
Use a sliding window with a frequency map to maintain the current window's character counts. For each right endpoint, expand the window and check if the frequencies are all equal. If not, shrink the window from the left until the condition is met or the window is empty.
Brute Force Approach
Iterate over all possible substrings using two nested loops. For each substring, count the frequency of each character and check if all non-zero frequencies are equal.
Verified Code Solutions
function solution(nums) {
let min = nums[0];
let max = nums[0];
for (let i = 1; i < nums.length; i++) {
if (nums[i] < min) {
min = nums[i];
} else if (nums[i] > max) {
max = nums[i];
}
}
return min + max;
}class Solution {
public:
int solution(vector<int>& nums) {
int min = nums[0];
int max = nums[0];
for (int i = 1; i < nums.size(); i++) {
if (nums[i] < min) {
min = nums[i];
} else if (nums[i] > max) {
max = nums[i];
}
}
return min + max;
}
};class Solution {
public int solution(int[] nums) {
int min = nums[0];
int max = nums[0];
for (int i = 1; i < nums.length; i++) {
if (nums[i] < min) {
min = nums[i];
} else if (nums[i] > max) {
max = nums[i];
}
}
return min + max;
}
}def solution(nums):
min_val = nums[0]
max_val = nums[0]
for num in nums[1:]:
if num < min_val:
min_val = num
elif num > max_val:
max_val = num
return min_val + max_valfunction solution(nums) {
let min = nums[0];
let max = nums[0];
for (let i = 1; i < nums.length; i++) {
if (nums[i] < min) {
min = nums[i];
} else if (nums[i] > max) {
max = nums[i];
}
}
return min + max;
}Asked in Top Tech Interviews
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