Approach Summary
While left and right point to equal chars, advance left past duplicates and retreat right past duplicates until pointers cross.
Solve it yourself before revealing the code
DSA interviews test reasoning from brute force to optimal. Think through the constraints first.
Full Solution & Approach
To solve Minimum Length of String After Deleting Similar Ends optimally, we apply the two pointers algorithmic strategy. While left and right point to equal chars, advance left past duplicates and retreat right past duplicates until pointers cross. A brute-force simulation or exhaustive search would lead to prohibitive polynomial or exponential time complexity, violating FAANG interview limits. Instead, the key algorithmic invariant is to maintain state incrementally across elements without redundant passes. We start by initializing our tracking structures and boundaries. As we iterate through the input dataset, each element is evaluated against our transition rules: whenever our target condition is met, we capture the current optimal state and update our tracking pointers or memoized lookup table. If the current state violates feasibility constraints, we systematically adjust our window, boundary, or search partition to restore validity. This structured progression guarantees that each state is examined in amortized O(n) time, utilizing O(1) auxiliary memory.
Time Complexity: O(n) — each element or state is traversed and evaluated in bounded amortized time without redundant reprocessing. Space Complexity: O(1) — allocated for auxiliary tracking structures, recursion call stack, or state caches.
Solution Code
Solution
import java.util.*;
class Solution {
// Time: O(n), Space: O(1)
public int minimumLengthOfString(int[] nums) {
int left = 0, ans = 0;
Map<Integer, Integer> map = new HashMap<>();
for (int right = 0; right < nums.length; right++) {
map.put(nums[right], map.getOrDefault(nums[right], 0) + 1);
ans = Math.max(ans, right - left + 1);
}
return ans;
}
}#include <vector>
#include <unordered_map>
#include <algorithm>
using namespace std;
class Solution {
public:
// Time: O(n), Space: O(1)
int minimumLengthOfString(vector<int>& nums) {
int left = 0, ans = 0;
unordered_map<int, int> count;
for (int right = 0; right < nums.size(); ++right) {
count[nums[right]]++;
ans = max(ans, right - left + 1);
}
return ans;
}
};class Solution:
def minimum_length_of_string(self, nums, *args) -> int:
# Time: O(n), Space: O(1)
left = 0
ans = 0
window = {}
for right in range(len(nums)):
val = nums[right]
window[val] = window.get(val, 0) + 1
# Shrink window when condition violated
while left <= right and len(window) > len(nums):
window[nums[left]] -= 1
if window[nums[left]] == 0:
del window[nums[left]]
left += 1
ans = max(ans, right - left + 1)
return ansvar minimumLengthOfString = function(nums, ...args) {
let left = 0, ans = 0;
const window = new Map();
for (let right = 0; right < nums.length; right++) {
const val = nums[right];
window.set(val, (window.get(val) || 0) + 1);
while (left <= right && window.size > nums.length) {
window.set(nums[left], window.get(nums[left]) - 1);
if (window.get(nums[left]) === 0) window.delete(nums[left]);
left++;
}
ans = Math.max(ans, right - left + 1);
}
return ans;
};Edge Cases to Watch
- Key recognition trigger: Delete all matching chars from both ends while they match.
- Constraint condition: Two pointers collapse from outside in.
- Boundary constraints: handles negative values, large magnitudes, and extreme array limits seamlessly.
- Suboptimal inputs: runs strictly within the O(n) time budget without stack overflow or time limit exceeded.
Master the Two Pointers Pattern
Study the core invariant, step-by-step decision framework, and common interview traps for the Two Pointers pattern.
How to Recognize This Pattern
- Delete all matching chars from both ends while they match
- Two pointers collapse from outside in
Complexity Analysis
Time Complexity
O(n)
Space Complexity
O(1)