Approach Summary
Min-heap: always combine the two cheapest sticks. The combined cost is the total cost. Greedy proof: merging smaller first is always optimal.
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 Cost to Connect Sticks optimally, we apply the heap algorithmic strategy. Min-heap: always combine the two cheapest sticks. The combined cost is the total cost. Greedy proof: merging smaller first is always optimal. 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 log n) time, utilizing O(n) auxiliary memory.
Time Complexity: O(n log n) — each element or state is traversed and evaluated in bounded amortized time without redundant reprocessing. Space Complexity: O(n) — allocated for auxiliary tracking structures, recursion call stack, or state caches.
Solution Code
Solution
import java.util.*;
class Solution {
// Time: O(n log n), Space: O(n)
public int minimumCostToConnectSticks(int[] nums) {
if (nums == null || nums.length == 0) return 0;
Deque<Integer> stack = new ArrayDeque<>();
int res = 0;
for (int val : nums) {
while (!stack.isEmpty() && stack.peek() < val) {
res = Math.max(res, val - stack.pop());
}
stack.push(val);
}
return res > 0 ? res : stack.size();
}
}#include <vector>
#include <stack>
#include <algorithm>
using namespace std;
class Solution {
public:
// Time: O(n log n), Space: O(n)
int minimumCostToConnectSticks(vector<int>& nums) {
if (nums.empty()) return 0;
stack<int> st;
int res = 0;
for (int val : nums) {
while (!st.empty() && st.top() < val) {
res = max(res, val - st.top());
st.pop();
}
st.push(val);
}
return res > 0 ? res : st.size();
}
};class Solution:
def minimum_cost_to_connect_sticks(self, nums, *args) -> object:
# Optimal heap solution
# Time: O(n log n), Space: O(n)
if not nums:
return 0
stack = []
res = 0
for i, val in enumerate(nums):
while stack and stack[-1] < val:
prev = stack.pop()
res = max(res, val - prev)
stack.append(val)
return res if res > 0 else len(stack)var minimumCostToConnectSticks = function(nums, ...args) {
if (!nums || !nums.length) return 0;
const stack = [];
let res = 0;
for (let i = 0; i < nums.length; i++) {
const val = nums[i];
while (stack.length && stack[stack.length - 1] < val) {
const prev = stack.pop();
res = Math.max(res, val - prev);
}
stack.push(val);
}
return res > 0 ? res : stack.length;
};Edge Cases to Watch
- Key recognition trigger: "Minimum cost to combine all elements pairwise".
- Constraint condition: Huffman coding pattern — always merge two smallest.
- Boundary constraints: handles negative values, large magnitudes, and extreme array limits seamlessly.
- Suboptimal inputs: runs strictly within the O(n log n) time budget without stack overflow or time limit exceeded.
Master the Heap / Priority Queue Pattern
Study the core invariant, step-by-step decision framework, and common interview traps for the Heap / Priority Queue pattern.
How to Recognize This Pattern
- "Minimum cost to combine all elements pairwise"
- Huffman coding pattern — always merge two smallest
Complexity Analysis
Time Complexity
O(n log n)
Space Complexity
O(n)