**Sliding Window Max - Amazon Top Interview Questions**

### Problem Statement :

Given a list of integers nums and an integer k, return the maximum values of each sublist of length k. Constraints 1 ≤ n ≤ 100,000 where n is the length of nums. 1 ≤ k ≤ 100,000. Example 1 Input nums = [10, 5, 2, 7, 8, 7] k = 3 Output [10, 7, 8, 8] Explanation 10 = max(10, 5, 2) 7 = max(5, 2, 7) 8 = max(2, 7, 8) 8 = max(7, 8, 7) Example 2 Input nums = [1, 2, 3, 4, 5, 4, 3, 2, 1] k = 3 Output [3, 4, 5, 5, 5, 4, 3] Example 3 Input nums = [3, 2, 1, 2, 3] k = 2 Output [3, 2, 2, 3] Example 4 Input nums = [3, 2, 1, 2, 3] k = 5 Output [3]

### Solution :

` ````
Solution in C++ :
vector<int> solve(vector<int>& nums, int k) {
deque<int> dq;
vector<int> ans;
for (int i = 0; i < nums.size(); i++) {
while (!dq.empty() and dq.front() <= i - k) dq.pop_front();
while (!dq.empty() and nums[dq.back()] <= nums[i]) dq.pop_back();
dq.push_back(i);
if (i >= k - 1) {
ans.push_back(nums[dq.front()]);
}
}
return ans;
}
```

` ````
Solution in Java :
import java.util.*;
class Solution {
public int[] solve(int[] nums, int k) {
int n = nums.length;
int[] res = new int[n - k + 1];
LinkedList<Integer> q = new LinkedList();
// first window
for (int i = 0; i < k; i++) {
while (q.size() > 0 && nums[i] > nums[q.getLast()]) {
q.removeLast();
}
q.addLast(i);
}
res[0] = nums[q.getFirst()];
for (int i = 1; i < n; i++) {
if (i + k - 1 >= n)
break;
while (q.size() > 0 && q.getFirst() < i) q.removeFirst();
while (q.size() > 0 && nums[i + k - 1] > nums[q.getLast()]) q.removeLast();
q.addLast(i + k - 1);
res[i] = nums[q.getFirst()];
}
return res;
}
}
```

` ````
Solution in Python :
class Solution:
def getMax_Pos(self, nums, l, h):
maxNo = nums[l]
maxPos = l
for i in range(l, h + 1):
if nums[i] > maxNo:
maxNo = nums[i]
maxPos = i
return (maxNo, maxPos)
def solve(self, nums, k):
n = len(nums)
maxNo, maxPos = self.getMax_Pos(nums, 0, k - 1)
l = 1
h = k
ans = []
ans.append(maxNo)
while h < n:
if maxPos < l:
maxNo, maxPos = self.getMax_Pos(nums, l, h)
elif nums[h] > maxNo:
maxNo, maxPos = nums[h], h
ans.append(maxNo)
l += 1
h += 1
return ans
```

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