Problem Statement:
Given an integer array nums
, return true
if there exists a triple of indices (i, j, k)
such that i < j < k
and nums[i] < nums[j] < nums[k]
. If no such indices exist, return false
.
Example 1:
- Input:
nums = [1,2,3,4,5]
- Output:
true
- Explanation: Any triplet where
i < j < k
is valid.
Example 2:
- Input:
nums = [5,4,3,2,1]
- Output:
false
- Explanation: No triplet exists.
Example 3:
- Input:
nums = [2,1,5,0,4,6]
- Output:
true
- Explanation: The triplet
(3, 4, 5)
is valid becausenums[3] == 0 < nums[4] == 4 < nums[5] == 6
.
Constraints:
1 <= nums.length <= 5 * 10^5
-2^31 <= nums[i] <= 2^31 - 1
Follow-up:
Can you implement a solution that runs in O(n) time complexity and O(1) space complexity?
Initial Thought Process:
To solve this problem efficiently, we need to keep track of the smallest and second smallest values encountered so far. If we find a third value that is greater than the second smallest, then we have found an increasing triplet.
Basic Solution (Brute Force):
The brute force solution involves checking all possible triplets to see if there exists one that satisfies the condition i < j < k
and nums[i] < nums[j] < nums[k]
. This approach has a time complexity of O(n^3), which is not efficient for large input sizes.
Code:
function increasingTripletBruteForce(nums: number[]): boolean {
const n = nums.length;
for (let i = 0; i < n - 2; i++) {
for (let j = i + 1; j < n - 1; j++) {
for (let k = j + 1; k < n; k++) {
if (nums[i] < nums[j] && nums[j] < nums[k]) {
return true;
}
}
}
}
return false;
}
Time Complexity Analysis:
- Time Complexity: O(n^3), where n is the length of the array. This is because we are checking all possible triplets.
- Space Complexity: O(1), as we are not using any extra space.
Limitations:
The brute force solution is not efficient and is not suitable for large input sizes.
Optimized Solution:
The optimized solution involves iterating through the array while maintaining two variables, first
and second
, which represent the smallest and second smallest values encountered so far. If we find a value greater than second
, then we return true
.
Code:
function increasingTriplet(nums: number[]): boolean {
let first = Infinity;
let second = Infinity;
for (let num of nums) {
if (num <= first) {
first = num; // smallest value
} else if (num <= second) {
second = num; // second smallest value
} else {
return true; // found a value greater than second smallest, thus an increasing triplet exists
}
}
return false;
}
Time Complexity Analysis:
- Time Complexity: O(n), where n is the length of the array. We iterate through the array once.
- Space Complexity: O(1), as we are using only a constant amount of extra space.
Improvements Over Basic Solution:
- This solution runs in linear time and uses constant space, making it optimal for the given constraints.
Edge Cases and Testing:
Edge Cases:
- The array is in decreasing order.
- The array contains exactly three elements in increasing order.
- The array has a large number of elements with no increasing triplet.
- The array contains duplicates.
Test Cases:
console.log(increasingTripletBruteForce([1,2,3,4,5])); // true
console.log(increasingTripletBruteForce([5,4,3,2,1])); // false
console.log(increasingTripletBruteForce([2,1,5,0,4,6])); // true
console.log(increasingTripletBruteForce([1,1,1,1,1])); // false
console.log(increasingTripletBruteForce([1,2])); // false
console.log(increasingTripletBruteForce([1,2,3])); // true
console.log(increasingTripletBruteForce([1,5,0,4,1,3])); // true
console.log(increasingTriplet([1,2,3,4,5])); // true
console.log(increasingTriplet([5,4,3,2,1])); // false
console.log(increasingTriplet([2,1,5,0,4,6])); // true
console.log(increasingTriplet([1,1,1,1,1])); // false
console.log(increasingTriplet([1,2])); // false
console.log(increasingTriplet([1,2,3])); // true
console.log(increasingTriplet([1,5,0,4,1,3])); // true
General Problem-Solving Strategies:
- Understand the Problem: Carefully read the problem statement to understand the requirements and constraints.
- Identify Key Operations: Determine the key operations needed, such as tracking the smallest and second smallest values.
- Optimize for Efficiency: Use efficient algorithms and data structures to minimize time and space complexity.
- Test Thoroughly: Test the solution with various cases, including edge cases, to ensure correctness.
Identifying Similar Problems:
-
Subarray Problems:
- Problems where you need to find subarrays with specific properties.
- Example: Finding the maximum sum subarray (Kadane's Algorithm).
-
Two-Pointer Technique:
- Problems where using two pointers can help optimize the solution.
- Example: Removing duplicates from a sorted array.
-
In-Place Algorithms:
- Problems where operations need to be performed in place with limited extra space.
- Example: Rotating an array to the right by k steps.
Conclusion:
- The problem of finding an increasing triplet subsequence can be efficiently solved using both a brute force approach and an optimized solution with linear time and constant space complexity.
- Understanding the problem and breaking it down into manageable parts is crucial.
- Using efficient algorithms ensures the solution is optimal for large inputs.
- Testing with various edge cases ensures robustness.
- Recognizing patterns in problems can help apply similar solutions to other challenges.
By practicing such problems and strategies, you can improve your problem-solving skills and be better prepared for various coding challenges.
Top comments (0)