**Shortest Sublist to Remove to Make Sorted List - Amazon Top Interview Questions**

### Problem Statement :

Given a list of integers nums, return the length of the shortest sublist that you can remove such that the resulting list is in ascending order. Constraints n ≤ 100,000 where n is the length of nums Example 1 Input nums = [1, 1, 3, 5, 4, 2, 1, 9] Output 3 Explanation If we remove [4, 2, 1], then the list would be sorted in ascending order: [1, 1, 3, 5, 9].

### Solution :

` ````
Solution in C++ :
int solve(vector<int>& nums) {
{
bool issorted = true;
for (int i = 1; i < nums.size() && issorted; i++) {
issorted = nums[i] >= nums[i - 1];
}
if (issorted) return 0;
}
int prefixsize = 1;
for (int i = 1; i < nums.size(); i++) {
if (nums[i] < nums[i - 1]) break;
prefixsize++;
}
int suffixsize = 1;
for (int i = nums.size() - 2; i >= 0; i--) {
if (nums[i] > nums[i + 1]) break;
suffixsize++;
}
int ret = suffixsize;
int j = 0;
for (int i = 0; i < prefixsize; i++) {
while (j < suffixsize && nums[nums.size() - suffixsize + j] < nums[i]) j++;
ret = max(ret, i + 1 + suffixsize - j);
}
return nums.size() - ret;
}
```

` ````
Solution in Java :
import java.util.*;
class Solution {
public int solve(int[] nums) {
final int N = nums.length;
if (N < 2)
return 0;
boolean[] leftvalid = new boolean[N];
leftvalid[0] = true;
for (int i = 1; i != N; i++) leftvalid[i] = leftvalid[i - 1] && (nums[i] >= nums[i - 1]);
boolean[] rightvalid = new boolean[N];
rightvalid[N - 1] = true;
for (int i = N - 2; i != -1; i--)
rightvalid[i] = rightvalid[i + 1] && (nums[i] <= nums[i + 1]);
if (rightvalid[0])
return 0;
int l = 1, h = N - 2;
while (l <= h) {
int m = (l + h) >>> 1;
if (can(nums, leftvalid, rightvalid, m))
h = m - 1;
else
l = m + 1;
}
return l;
}
private boolean can(int[] nums, boolean[] left, boolean[] right, final int K) {
if (left[left.length - K - 1] || right[K])
return true;
for (int i = 1, j = K; j + 1 != left.length; i++, j++)
if (left[i - 1] && right[j + 1] && nums[i - 1] <= nums[j + 1])
return true;
return false;
}
}
```

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