Make Array Empty - Practice Coding Problems

Make Array Empty - Problem

Array Binary Search Greedy Binary Indexed Tree Segment Tree Sorting Ordered Set Hard

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.

Input & Output

Example 1 — Basic Case

$ Input: nums = [3,1,4,2]

› Output: 7

💡 Note: Operations: [3,1,4,2]→[1,4,2,3]→[4,2,3]→[2,3,4]→[3,4]→[4,3]→[4]→[]. Total: 7 operations.

Example 2 — Already Sorted

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

› Output: 3

💡 Note: Elements are already in ascending order, so each can be removed immediately: 1+1+1 = 3 operations.

Example 3 — Reverse Order

$ Input: nums = [4,3,2,1]

› Output: 10

💡 Note: Worst case: need to rotate many times. Remove 1 (pos 3): 4 ops. Remove 2 (pos 2): 3 ops. Remove 3 (pos 1): 2 ops. Remove 4 (pos 0): 1 op. Total: 10.

Constraints

1 ≤ nums.length ≤ 10⁵
1 ≤ nums[i] ≤ 10⁹
All elements in nums are distinct

Visualization

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Understanding the Visualization

Input Array

[3,1,4,2] - mixed order elements

Operation Rules

Remove if first is minimum, else rotate

Result

7 total operations needed

Key Takeaway

🎯 Key Insight: Elements must be removed in ascending order, allowing mathematical calculation instead of simulation

Asked in

G Google 15 M Microsoft 12 f Meta 8

The key insight is that elements must be removed in ascending order, so we can calculate operations mathematically. The optimal approach sorts the array with indices and computes the position adjustments. Time: O(n log n), Space: O(n)

Common Approaches

Approach	Time	Space	Notes
✓ Mathematical Formula	O(n log n)	O(n)	Calculate operations using position differences without simulation
Sort-Based Solution	O(n²)	O(n)	Pre-sort to determine removal order, then calculate positions
Simulation	O(n²)	O(1)	Simulate the process step by step

Mathematical Formula — Algorithm Steps

Sort array with original indices
For each element, count how many smaller elements come after it
Use formula: position + count of smaller elements after
Sum all operations

Visualization

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Step-by-Step Walkthrough

Create Pairs

Store (value, original_index) pairs

Sort by Value

Sort to get removal order

Calculate Formula

For each element: current_position + 1

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

typedef struct {
    int value;
    int index;
} Pair;

int compare(const void* a, const void* b) {
    Pair* pa = (Pair*)a;
    Pair* pb = (Pair*)b;
    return pa->value - pb->value;
}

int solution(int* nums, int size) {
    Pair* pairs = (Pair*)malloc(size * sizeof(Pair));
    
    for (int i = 0; i < size; i++) {
        pairs[i].value = nums[i];
        pairs[i].index = i;
    }
    
    qsort(pairs, size, sizeof(Pair), compare);
    
    int operations = 0;
    int removed = 0;
    
    for (int i = 0; i < size; i++) {
        int currentPos = pairs[i].index - removed;
        operations += currentPos + 1;
        removed++;
    }
    
    free(pairs);
    return operations;
}

int main() {
    char line[1000];
    fgets(line, sizeof(line), stdin);
    
    int nums[100];
    int size = 0;
    
    char* token = strtok(line, "[],\n ");
    while (token != NULL) {
        nums[size++] = atoi(token);
        token = strtok(NULL, "[],\n ");
    }
    
    printf("%d\n", solution(nums, size));
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n log n)

Sorting takes O(n log n), then single pass to calculate operations

⚡ Linearithmic

Space Complexity

O(n)

Extra space for storing value-index pairs

⚡ Linearithmic Space

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Smart Actions

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Mathematical Formula — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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