LeetCode #2659 — HARD

Make Array Empty

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

The Problem

Problem Statement

You are given an integer array nums containing distinct numbers, and you can perform the following operations until the array is empty:

If the first element has the smallest value, remove it
Otherwise, put the first element at the end of the array.

Return an integer denoting the number of operations it takes to make nums empty.

Example 1:

Input: nums = [3,4,-1]
Output: 5

Operation	Array
1	[4, -1, 3]
2	[-1, 3, 4]
3	[3, 4]
4	[4]
5	[]

Example 2:

Input: nums = [1,2,4,3]
Output: 5

Operation	Array
1	[2, 4, 3]
2	[4, 3]
3	[3, 4]
4	[4]
5	[]

Example 3:

Input: nums = [1,2,3]
Output: 3

Operation	Array
1	[2, 3]
2	[3]
3	[]

Constraints:

1 <= nums.length <= 10⁵
-10⁹<= nums[i] <= 10⁹
All values in nums are distinct.

Step 01

Brute Force Baseline

Problem summary: You are given an integer array nums containing distinct numbers, and you can perform the following operations until the array is empty: If the first element has the smallest value, remove it Otherwise, put the first element at the end of the array. Return an integer denoting the number of operations it takes to make nums empty.

Baseline thinking

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

Pattern signal: Array · Binary Search · Greedy · Segment Tree

Example 1

[3,4,-1]

Example 2

[1,2,4,3]

Example 3

[1,2,3]

Step 02

Core Insight

What unlocks the optimal approach

Understand the order in which the indices are removed from the array.
We don’t really need to delete or move the elements, only the array length matters.
Upon removing an index, decide how many steps it takes to move to the next one.
Use a data structure to speed up the calculation.

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 #2659: Make Array Empty
class BinaryIndexedTree {
    private int n;
    private int[] c;

    public BinaryIndexedTree(int n) {
        this.n = n;
        c = new int[n + 1];
    }

    public void update(int x, int delta) {
        while (x <= n) {
            c[x] += delta;
            x += x & -x;
        }
    }

    public int query(int x) {
        int s = 0;
        while (x > 0) {
            s += c[x];
            x -= x & -x;
        }
        return s;
    }
}

class Solution {
    public long countOperationsToEmptyArray(int[] nums) {
        int n = nums.length;
        Map<Integer, Integer> pos = new HashMap<>();
        for (int i = 0; i < n; ++i) {
            pos.put(nums[i], i);
        }
        Arrays.sort(nums);
        long ans = pos.get(nums[0]) + 1;
        BinaryIndexedTree tree = new BinaryIndexedTree(n);
        for (int k = 0; k < n - 1; ++k) {
            int i = pos.get(nums[k]), j = pos.get(nums[k + 1]);
            long d = j - i - (tree.query(j + 1) - tree.query(i + 1));
            ans += d + (n - k) * (i > j ? 1 : 0);
            tree.update(i + 1, 1);
        }
        return ans;
    }
}

// Accepted solution for LeetCode #2659: Make Array Empty
type BinaryIndexedTree struct {
	n int
	c []int
}

func newBinaryIndexedTree(n int) *BinaryIndexedTree {
	c := make([]int, n+1)
	return &BinaryIndexedTree{n, c}
}

func (this *BinaryIndexedTree) update(x, delta int) {
	for x <= this.n {
		this.c[x] += delta
		x += x & -x
	}
}

func (this *BinaryIndexedTree) query(x int) int {
	s := 0
	for x > 0 {
		s += this.c[x]
		x -= x & -x
	}
	return s
}

func countOperationsToEmptyArray(nums []int) int64 {
	n := len(nums)
	pos := map[int]int{}
	for i, x := range nums {
		pos[x] = i
	}
	sort.Ints(nums)
	tree := newBinaryIndexedTree(n)
	ans := pos[nums[0]] + 1
	for k := 0; k < n-1; k++ {
		i, j := pos[nums[k]], pos[nums[k+1]]
		d := j - i - (tree.query(j+1) - tree.query(i+1))
		if i > j {
			d += n - k
		}
		ans += d
		tree.update(i+1, 1)
	}
	return int64(ans)
}

# Accepted solution for LeetCode #2659: Make Array Empty
class Solution:
    def countOperationsToEmptyArray(self, nums: List[int]) -> int:
        pos = {x: i for i, x in enumerate(nums)}
        nums.sort()
        sl = SortedList()
        ans = pos[nums[0]] + 1
        n = len(nums)
        for k, (a, b) in enumerate(pairwise(nums)):
            i, j = pos[a], pos[b]
            d = j - i - sl.bisect(j) + sl.bisect(i)
            ans += d + (n - k) * int(i > j)
            sl.add(i)
        return ans

// Accepted solution for LeetCode #2659: Make Array Empty
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #2659: Make Array Empty
// class BinaryIndexedTree {
//     private int n;
//     private int[] c;
// 
//     public BinaryIndexedTree(int n) {
//         this.n = n;
//         c = new int[n + 1];
//     }
// 
//     public void update(int x, int delta) {
//         while (x <= n) {
//             c[x] += delta;
//             x += x & -x;
//         }
//     }
// 
//     public int query(int x) {
//         int s = 0;
//         while (x > 0) {
//             s += c[x];
//             x -= x & -x;
//         }
//         return s;
//     }
// }
// 
// class Solution {
//     public long countOperationsToEmptyArray(int[] nums) {
//         int n = nums.length;
//         Map<Integer, Integer> pos = new HashMap<>();
//         for (int i = 0; i < n; ++i) {
//             pos.put(nums[i], i);
//         }
//         Arrays.sort(nums);
//         long ans = pos.get(nums[0]) + 1;
//         BinaryIndexedTree tree = new BinaryIndexedTree(n);
//         for (int k = 0; k < n - 1; ++k) {
//             int i = pos.get(nums[k]), j = pos.get(nums[k + 1]);
//             long d = j - i - (tree.query(j + 1) - tree.query(i + 1));
//             ans += d + (n - k) * (i > j ? 1 : 0);
//             tree.update(i + 1, 1);
//         }
//         return ans;
//     }
// }

// Accepted solution for LeetCode #2659: Make Array Empty
class BinaryIndexedTree {
    private n: number;
    private c: number[];

    constructor(n: number) {
        this.n = n;
        this.c = Array(n + 1).fill(0);
    }

    public update(x: number, v: number): void {
        while (x <= this.n) {
            this.c[x] += v;
            x += x & -x;
        }
    }

    public query(x: number): number {
        let s = 0;
        while (x > 0) {
            s += this.c[x];
            x -= x & -x;
        }
        return s;
    }
}

function countOperationsToEmptyArray(nums: number[]): number {
    const pos: Map<number, number> = new Map();
    const n = nums.length;
    for (let i = 0; i < n; ++i) {
        pos.set(nums[i], i);
    }
    nums.sort((a, b) => a - b);
    const tree = new BinaryIndexedTree(n);
    let ans = pos.get(nums[0])! + 1;
    for (let k = 0; k < n - 1; ++k) {
        const i = pos.get(nums[k])!;
        const j = pos.get(nums[k + 1])!;
        let d = j - i - (tree.query(j + 1) - tree.query(i + 1));
        if (i > j) {
            d += n - k;
        }
        ans += d;
        tree.update(i + 1, 1);
    }
    return ans;
}

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(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.

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.

Using greedy without proof

Wrong move: Locally optimal choices may fail globally.

Usually fails on: Counterexamples appear on crafted input orderings.

Fix: Verify with exchange argument or monotonic objective before committing.