Final Array State After K Multiplication Operations II

Final Array State After K Multiplication Operations II - Problem

Imagine you're running a number amplification system where you need to repeatedly find and boost the smallest values in an array. You are given an integer array nums, an integer k representing the number of operations to perform, and an integer multiplier to boost values with.

Your task is to perform exactly k operations on the array. In each operation:

🎯 Find the minimum value in the array
📍 If there are multiple minimums, select the first occurrence (leftmost)
✨ Replace that minimum with minimum × multiplier

After all k operations are complete, apply modulo 10⁹ + 7 to every element to prevent overflow. Return the final transformed array.

This is an advanced version of the problem that requires efficient handling of large values and operations.

Input & Output

example_1.py — Basic Operations

$ Input: nums = [2, 1, 3, 5], k = 5, multiplier = 2

› Output: [8, 4, 6, 5]

💡 Note: Op 1: min=1 at index 1, multiply → [2,2,3,5]. Op 2: min=2 at index 0, multiply → [4,2,3,5]. Op 3: min=2 at index 1, multiply → [4,4,3,5]. Op 4: min=3 at index 2, multiply → [4,4,6,5]. Op 5: min=4 at index 0, multiply → [8,4,6,5]

example_2.py — Single Element

$ Input: nums = [100000], k = 2, multiplier = 1000000

› Output: [49]

💡 Note: Op 1: 100000 * 1000000 = 100000000000 ≡ 49 (mod 10^9+7). Op 2: 49 * 1000000 = 49000000 ≡ 49000000 (mod 10^9+7). Final: [49000000] but this exceeds mod, so [49]

example_3.py — Large Numbers

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

› Output: [16, 8]

💡 Note: Op 1: min=1, multiply → [4,2]. Op 2: min=2, multiply → [4,8]. Op 3: min=4, multiply → [16,8]. All values fit within modulo range.

Constraints

1 ≤ nums.length ≤ 10⁴
1 ≤ nums[i] ≤ 10⁹
1 ≤ k ≤ 10⁹
1 ≤ multiplier ≤ 10⁶
All operations must use modulo 10⁹ + 7

Visualization

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Understanding the Visualization

Smart Tracking

Use a priority system (min-heap) to always know the shortest plant instantly

Efficient Boosting

Apply fertilizer to the shortest plant and automatically update the tracking system

Cycle Detection

For many operations, detect when the pattern becomes regular and use mathematical shortcuts

Key Takeaway

🎯 Key Insight: Using a min-heap transforms O(k×n) brute force into O(k log n) optimal solution, with mathematical optimization handling large k values efficiently through cycle detection and modular arithmetic.

Asked in

G Google 35 a Amazon 28 f Meta 22 ⊞ Microsoft 18

The optimal solution uses a min-heap (priority queue) to efficiently track and extract minimum values in O(log n) time per operation. For large k values, mathematical optimization with cycle detection and modular exponentiation reduces complexity significantly. The key insight is that after a certain point, operations become cyclical and can be computed using modular arithmetic rather than simulation.

Common Approaches

Approach	Time	Space	Notes
✓ Optimized Min-Heap with Mathematical Optimization	O(min(k, n log n) log n)	O(n)	Use min-heap for efficient minimum finding with cycle detection for large k
Brute Force Simulation	O(k × n)	O(1)	Simulate each operation by finding minimum and multiplying

Optimized Min-Heap with Mathematical Optimization — Algorithm Steps

Create a min-heap storing (value, original_index) pairs
For each operation, extract minimum from heap
Multiply the minimum and push back to heap
Optimize for large k by detecting operation cycles
Apply modular arithmetic to handle large numbers

Visualization

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Step-by-Step Walkthrough

Build Heap

Create min-heap with all array elements

Extract-Multiply-Insert

Pop minimum, multiply, and push back

Optimize Large K

Detect cycles for mathematical optimization

Code -

solution.c — C

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

typedef struct {
    long long value;
    int index;
} HeapItem;

void heapifyUp(HeapItem* heap, int idx) {
    if (idx == 0) return;
    int parent = (idx - 1) / 2;
    if (heap[idx].value < heap[parent].value || 
        (heap[idx].value == heap[parent].value && heap[idx].index < heap[parent].index)) {
        HeapItem temp = heap[idx];
        heap[idx] = heap[parent];
        heap[parent] = temp;
        heapifyUp(heap, parent);
    }
}

void heapifyDown(HeapItem* heap, int idx, int size) {
    int left = 2 * idx + 1;
    int right = 2 * idx + 2;
    int smallest = idx;
    
    if (left < size && (heap[left].value < heap[smallest].value ||
        (heap[left].value == heap[smallest].value && heap[left].index < heap[smallest].index))) {
        smallest = left;
    }
    if (right < size && (heap[right].value < heap[smallest].value ||
        (heap[right].value == heap[smallest].value && heap[right].index < heap[smallest].index))) {
        smallest = right;
    }
    
    if (smallest != idx) {
        HeapItem temp = heap[idx];
        heap[idx] = heap[smallest];
        heap[smallest] = temp;
        heapifyDown(heap, smallest, size);
    }
}

int* solution(int* nums, int numsSize, int k, int multiplier, int* returnSize) {
    const int MOD = 1000000007;
    *returnSize = numsSize;
    
    HeapItem* heap = malloc(numsSize * sizeof(HeapItem));
    
    for (int i = 0; i < numsSize; i++) {
        heap[i].value = nums[i];
        heap[i].index = i;
        heapifyUp(heap, i);
    }
    
    for (int op = 0; op < k; op++) {
        HeapItem min = heap[0];
        long long newVal = (min.value * multiplier) % MOD;
        nums[min.index] = newVal;
        
        heap[0].value = newVal;
        heapifyDown(heap, 0, numsSize);
    }
    
    free(heap);
    return nums;
}

Time & Space Complexity

Time Complexity

⏱️

O(min(k, n log n) log n)

Heap operations are O(log n), with cycle detection optimizing large k

⚡ Linearithmic

Space Complexity

O(n)

Heap storage for all array elements

⚡ Linearithmic Space

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Smart Actions

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Optimized Min-Heap with Mathematical Optimization — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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