LeetCode #2009 — HARD

Minimum Number of Operations to Make Array Continuous

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

Solve on LeetCode

The Problem

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 <= 10⁵
1 <= nums[i] <= 10⁹

Roadmap

Brute Force Baseline
Core Insight
Algorithm Walkthrough
Edge Cases
Full Annotated Code
Interactive Study Demo
Complexity Analysis

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Hash Map · Binary Search · Sliding Window

Example 1

[4,2,5,3]

Example 2

[1,2,3,5,6]

Example 3

[1,10,100,1000]

Related Problems

Longest Repeating Character Replacement (longest-repeating-character-replacement)
Continuous Subarray Sum (continuous-subarray-sum)
Moving Stones Until Consecutive II (moving-stones-until-consecutive-ii)
Minimum One Bit Operations to Make Integers Zero (minimum-one-bit-operations-to-make-integers-zero)
Minimum Adjacent Swaps for K Consecutive Ones (minimum-adjacent-swaps-for-k-consecutive-ones)

Step 02

Core Insight

What unlocks the optimal approach

Sort the array.
For every index do a binary search to get the possible right end of the window and calculate the possible answer.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Largest constraint values

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2009: Minimum Number of Operations to Make Array Continuous
class Solution {
    public int minOperations(int[] nums) {
        int n = nums.length;
        Arrays.sort(nums);
        int m = 1;
        for (int i = 1; i < n; ++i) {
            if (nums[i] != nums[i - 1]) {
                nums[m++] = nums[i];
            }
        }
        int ans = n;
        for (int i = 0; i < m; ++i) {
            int j = search(nums, nums[i] + n - 1, i, m);
            ans = Math.min(ans, n - (j - i));
        }
        return ans;
    }

    private int search(int[] nums, int x, int left, int right) {
        while (left < right) {
            int mid = (left + right) >> 1;
            if (nums[mid] > x) {
                right = mid;
            } else {
                left = mid + 1;
            }
        }
        return left;
    }
}

// Accepted solution for LeetCode #2009: Minimum Number of Operations to Make Array Continuous
func minOperations(nums []int) int {
	sort.Ints(nums)
	n := len(nums)
	m := 1
	for i := 1; i < n; i++ {
		if nums[i] != nums[i-1] {
			nums[m] = nums[i]
			m++
		}
	}
	ans := n
	for i := 0; i < m; i++ {
		j := sort.Search(m, func(k int) bool { return nums[k] > nums[i]+n-1 })
		ans = min(ans, n-(j-i))
	}
	return ans
}

# Accepted solution for LeetCode #2009: Minimum Number of Operations to Make Array Continuous
class Solution:
    def minOperations(self, nums: List[int]) -> int:
        ans = n = len(nums)
        nums = sorted(set(nums))
        for i, v in enumerate(nums):
            j = bisect_right(nums, v + n - 1)
            ans = min(ans, n - (j - i))
        return ans

// Accepted solution for LeetCode #2009: Minimum Number of Operations to Make Array Continuous
use std::collections::BTreeSet;

impl Solution {
    #[allow(dead_code)]
    pub fn min_operations(nums: Vec<i32>) -> i32 {
        let n = nums.len();
        let nums = nums.into_iter().collect::<BTreeSet<i32>>();

        let m = nums.len();
        let nums = nums.into_iter().collect::<Vec<i32>>();

        let mut ans = n;

        for i in 0..m {
            let j = match nums.binary_search(&(nums[i] + (n as i32))) {
                Ok(idx) => idx,
                Err(idx) => idx,
            };
            ans = std::cmp::min(ans, n - (j - i));
        }

        ans as i32
    }
}

// Accepted solution for LeetCode #2009: Minimum Number of Operations to Make Array Continuous
function minOperations(nums: number[]): number {
    const n = nums.length;
    nums.sort((a, b) => a - b);
    let m = 1;
    for (let i = 1; i < n; ++i) {
        if (nums[i] !== nums[i - 1]) {
            nums[m++] = nums[i];
        }
    }
    let ans = n;
    for (let i = 0; i < m; ++i) {
        const j = search(nums, nums[i] + n - 1, i, m);
        ans = Math.min(ans, n - (j - i));
    }
    return ans;
}

function search(nums: number[], x: number, left: number, right: number): number {
    while (left < right) {
        const mid = (left + right) >> 1;
        if (nums[mid] > x) {
            right = mid;
        } else {
            left = mid + 1;
        }
    }
    return left;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n × log n)

Space

O(log n)

Approach Breakdown

LINEAR SCAN

O(n) time

O(1) space

Check every element from left to right until we find the target or exhaust the array. Each comparison is O(1), and we may visit all n elements, giving O(n). No extra space needed.

BINARY SEARCH

O(log n) time

O(1) space

Each comparison eliminates half the remaining search space. After k comparisons, the space is n/2ᵏ. We stop when the space is 1, so k = log₂ n. No extra memory needed — just two pointers (lo, hi).

Shortcut: Halving the input each step → O(log n). Works on any monotonic condition, not just sorted arrays.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.

Boundary update without `+1` / `-1`

Wrong move: Setting `lo = mid` or `hi = mid` can stall and create an infinite loop.

Usually fails on: Two-element ranges never converge.

Fix: Use `lo = mid + 1` or `hi = mid - 1` where appropriate.

Shrinking the window only once

Wrong move: Using `if` instead of `while` leaves the window invalid for multiple iterations.

Usually fails on: Over-limit windows stay invalid and produce wrong lengths/counts.

Fix: Shrink in a `while` loop until the invariant is valid again.