solutions: 2009 - Minimum Number of Operations to Make Array Continuo…

…us (Hard)

solutions/2000-2099/2009-minimum-number-of-operations-to-make-array-continuous-hard.md

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+---
+description: 'Author: @wingkwong | https://leetcode.com/problems/minimum-number-of-operations-to-make-array-continuous/'
+tags: [Array, Binary Search]
+---
+
+# 2009 - Minimum Number of Operations to Make Array Continuous (Hard) 
+
+## Problem Link
+
+https://leetcode.com/problems/minimum-number-of-operations-to-make-array-continuous/
+
+## Problem Statement
+
+You are given an integer array `nums`. In one operation, you can replace **any** element in `nums` with **any** integer.
+
+`nums` is considered **continuous** if both of the following conditions are fulfilled:
+
+- All elements in `nums` are **unique**.
+- The difference between the **maximum** element and the **minimum** element in `nums` equals `nums.length - 1`.
+
+For example, `nums = [4, 2, 5, 3]` is **continuous**, but `nums = [1, 2, 3, 5, 6]` is **not continuous**.
+
+Return *the **minimum** number of operations to make*`nums`**continuous**.
+
+**Example 1:**
+
+```
+Input: nums = [4,2,5,3]
+Output: 0
+Explanation: nums is already continuous.
+```
+
+**Example 2:**
+
+```
+Input: nums = [1,2,3,5,6]
+Output: 1
+Explanation: One possible solution is to change the last element to 4.
+The resulting array is [1,2,3,5,4], which is continuous.
+```
+
+**Example 3:**
+
+```
+Input: nums = [1,10,100,1000]
+Output: 3
+Explanation: One possible solution is to:
+- Change the second element to 2.
+- Change the third element to 3.
+- Change the fourth element to 4.
+The resulting array is [1,2,3,4], which is continuous.
+```
+
+**Constraints:**
+
+- `1 <= nums.length <= 1e5`
+- `1 <= nums[i] <= 1e9`
+
+## Approach 1: Sliding Window
+
+<Tabs>
+<TabItem value="kotlin" label="Kotlin">
+<SolutionAuthor name="@wingkwong"/>
+
+```kt
+class Solution {
+    fun minOperations(nums: IntArray): Int {
+        val n = nums.size
+        val v = nums.toSortedSet().toIntArray()
+        val m = v.size
+        var ans = n
+        var r = 0
+        for (l in 0 until m) {
+            // assuming v[l] is the min element in the continuous array
+            // the maximum element v[r] would be v[r] - v[l] + 1 = n -> v[r] = v[l] + n - 1
+            // find the target `r` / use binary search here
+            while (r < m && v[r] <= v[l] + n - 1) r++
+            // now we got r - l elements are within the range
+            // the rest of the elements `n - (r - l)` will be replaced
+            // i.e. replace all elements to v[l] in [0 .. l) and v[r] in [r .. m)
+            ans = min(ans, n - (r - l))
+        }        
+        return ans
+    }
+}
+```
+
+</TabItem>
+</Tabs>