LeetCode #3266 — HARD

Final Array State After K Multiplication Operations II

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, an integer k, and an integer multiplier.

You need to perform k operations on nums. In each operation:

Find the minimum value x in nums. If there are multiple occurrences of the minimum value, select the one that appears first.
Replace the selected minimum value x with x * multiplier.

After the k operations, apply modulo 10⁹ + 7 to every value in nums.

Return an integer array denoting the final state of nums after performing all k operations and then applying the modulo.

Example 1:

Input: nums = [2,1,3,5,6], k = 5, multiplier = 2

Output: [8,4,6,5,6]

Explanation:

Operation	Result
After operation 1	[2, 2, 3, 5, 6]
After operation 2	[4, 2, 3, 5, 6]
After operation 3	[4, 4, 3, 5, 6]
After operation 4	[4, 4, 6, 5, 6]
After operation 5	[8, 4, 6, 5, 6]
After applying modulo	[8, 4, 6, 5, 6]

Example 2:

Input: nums = [100000,2000], k = 2, multiplier = 1000000

Output: [999999307,999999993]

Explanation:

Operation	Result
After operation 1	[100000, 2000000000]
After operation 2	[100000000000, 2000000000]
After applying modulo	[999999307, 999999993]

Constraints:

1 <= nums.length <= 10⁴
1 <= nums[i] <= 10⁹
1 <= k <= 10⁹
1 <= multiplier <= 10⁶

Step 01

Brute Force Baseline

Problem summary: You are given an integer array nums, an integer k, and an integer multiplier. You need to perform k operations on nums. In each operation: Find the minimum value x in nums. If there are multiple occurrences of the minimum value, select the one that appears first. Replace the selected minimum value x with x * multiplier. After the k operations, apply modulo 109 + 7 to every value in nums. Return an integer array denoting the final state of nums after performing all k operations and then applying the modulo.

Baseline thinking

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

Pattern signal: Array

Example 1

[2,1,3,5,6]
5
2

Example 2

[100000,2000]
2
1000000

Step 02

Core Insight

What unlocks the optimal approach

What happens when <code>min(nums) * multiplier > max(nums)</code>?
A cycle of operations begins.
Simulate until <code>min(nums) * multiplier > max(nums)</code>, then greedily distribute remaining multiplications.

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 #3266: Final Array State After K Multiplication Operations II
class Solution {
    public int[] getFinalState(int[] nums, int k, int multiplier) {
        if (multiplier == 1) {
            return nums;
        }
        PriorityQueue<long[]> pq = new PriorityQueue<>(
            (a, b) -> a[0] == b[0] ? Long.compare(a[1], b[1]) : Long.compare(a[0], b[0]));
        int n = nums.length;
        int m = Arrays.stream(nums).max().getAsInt();
        for (int i = 0; i < n; ++i) {
            pq.offer(new long[] {nums[i], i});
        }
        for (; k > 0 && pq.peek()[0] < m; --k) {
            long[] p = pq.poll();
            p[0] *= multiplier;
            pq.offer(p);
        }
        final int mod = (int) 1e9 + 7;
        for (int i = 0; i < n; ++i) {
            long[] p = pq.poll();
            long x = p[0];
            int j = (int) p[1];
            nums[j] = (int) ((x % mod) * qpow(multiplier, k / n + (i < k % n ? 1 : 0), mod) % mod);
        }
        return nums;
    }

    private int qpow(long a, long n, long mod) {
        long ans = 1 % mod;
        for (; n > 0; n >>= 1) {
            if ((n & 1) == 1) {
                ans = ans * a % mod;
            }
            a = a * a % mod;
        }
        return (int) ans;
    }
}

// Accepted solution for LeetCode #3266: Final Array State After K Multiplication Operations II
func getFinalState(nums []int, k int, multiplier int) []int {
	if multiplier == 1 {
		return nums
	}
	n := len(nums)
	pq := make(hp, n)
	for i, x := range nums {
		pq[i] = pair{x, i}
	}
	heap.Init(&pq)
	m := slices.Max(nums)
	for ; k > 0 && pq[0].x < m; k-- {
		x := pq[0]
		heap.Pop(&pq)
		x.x *= multiplier
		heap.Push(&pq, x)
	}
	const mod int = 1e9 + 7

	for i := range nums {
		p := heap.Pop(&pq).(pair)
		x, j := p.x, p.i
		power := k / n
		if i < k%n {
			power++
		}
		nums[j] = (x % mod) * qpow(multiplier, power, mod) % mod
	}
	return nums
}

func qpow(a, n, mod int) int {
	ans := 1 % mod
	a = a % mod
	for n > 0 {
		if n&1 == 1 {
			ans = (ans * a) % mod
		}
		a = (a * a) % mod
		n >>= 1
	}
	return int(ans)
}

type pair struct{ x, i int }
type hp []pair

func (h hp) Len() int           { return len(h) }
func (h hp) Less(i, j int) bool { return h[i].x < h[j].x || h[i].x == h[j].x && h[i].i < h[j].i }
func (h hp) Swap(i, j int)      { h[i], h[j] = h[j], h[i] }
func (h *hp) Push(x any)        { *h = append(*h, x.(pair)) }
func (h *hp) Pop() (x any)      { a := *h; x = a[len(a)-1]; *h = a[:len(a)-1]; return x }

# Accepted solution for LeetCode #3266: Final Array State After K Multiplication Operations II
class Solution:
    def getFinalState(self, nums: List[int], k: int, multiplier: int) -> List[int]:
        if multiplier == 1:
            return nums
        pq = [(x, i) for i, x in enumerate(nums)]
        heapify(pq)
        m = max(nums)
        while k and pq[0][0] < m:
            x, i = heappop(pq)
            heappush(pq, (x * multiplier, i))
            k -= 1
        n = len(nums)
        mod = 10**9 + 7
        pq.sort()
        for i, (x, j) in enumerate(pq):
            nums[j] = x * pow(multiplier, k // n + int(i < k % n), mod) % mod
        return nums

// Accepted solution for LeetCode #3266: Final Array State After K Multiplication Operations II
/**
 * [3266] Final Array State After K Multiplication Operations II
 */
pub struct Solution {}

// submission codes start here
use std::cmp::Ordering;
use std::collections::BinaryHeap;

#[derive(Debug, Eq, PartialEq)]
struct Number {
    value: i64,
    pos: usize,
}

impl Ord for Number {
    fn cmp(&self, other: &Self) -> Ordering {
        match other.value.cmp(&self.value) {
            Ordering::Equal => other.pos.cmp(&self.pos),
            r => r,
        }
    }
}

impl PartialOrd for Number {
    fn partial_cmp(&self, other: &Self) -> Option<Ordering> {
        Some(self.cmp(other))
    }
}

const MOD: i64 = 1_000_000_007;

impl Solution {
    pub fn get_final_state(nums: Vec<i32>, k: i32, multiplier: i32) -> Vec<i32> {
        let multiplier = multiplier as i64;

        if multiplier == 1 {
            return nums;
        }

        let n = nums.len();
        let max_number = *nums.iter().max().unwrap() as i64;
        let mut heap = BinaryHeap::new();

        for (i, v) in nums.into_iter().enumerate() {
            heap.push(Number {
                value: v as i64,
                pos: i,
            });
        }

        let mut actual_k = k;
        for _ in 0..k {
            if let Some(head) = heap.peek() {
                if head.value >= max_number {
                    break;
                }
            }

            let head = heap.pop().unwrap();

            heap.push(Number {
                value: head.value * multiplier,
                pos: head.pos,
            });
            actual_k -= 1;
        }

        let mut result = vec![0; n];

        let k = actual_k as usize;
        // 牙医 shake it!
        // 为什么heap.iter()会是以随机的顺序返回！
        for (i, number) in std::iter::from_fn(move || heap.pop()).enumerate() {
            let t = k / n + if i < k % n { 1 } else { 0 };
            result[number.pos] =
                (number.value % MOD * Self::quick_mulitplx(multiplier, t as i64) % MOD) as i32;
        }

        result
    }

    fn quick_mulitplx(mut x: i64, mut y: i64) -> i64 {
        let mut result = 1;

        while y > 0 {
            if y & 1 == 1 {
                result = (result * x) % MOD;
            }

            y = y >> 1;
            x = (x * x) % MOD;
        }

        result
    }
}

// submission codes end

#[cfg(test)]
mod tests {
    use super::*;

    #[test]
    fn test_3266() {
        assert_eq!(
            vec![664480092, 763599523, 886046925, 47878852],
            Solution::get_final_state(
                vec![66307295, 441787703, 589039035, 322281864],
                900_900_704,
                641725
            )
        );
        assert_eq!(
            vec![8, 8, 4],
            Solution::get_final_state(vec![4, 2, 4], 3, 2)
        );
        assert_eq!(
            vec![8, 4, 6, 5, 6],
            Solution::get_final_state(vec![2, 1, 3, 5, 6], 5, 2)
        );
        assert_eq!(
            vec![999_999_307, 999_999_993],
            Solution::get_final_state(vec![100_000, 2_000], 2, 1_000_000)
        );
    }
}

// Accepted solution for LeetCode #3266: Final Array State After K Multiplication Operations II
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #3266: Final Array State After K Multiplication Operations II
// class Solution {
//     public int[] getFinalState(int[] nums, int k, int multiplier) {
//         if (multiplier == 1) {
//             return nums;
//         }
//         PriorityQueue<long[]> pq = new PriorityQueue<>(
//             (a, b) -> a[0] == b[0] ? Long.compare(a[1], b[1]) : Long.compare(a[0], b[0]));
//         int n = nums.length;
//         int m = Arrays.stream(nums).max().getAsInt();
//         for (int i = 0; i < n; ++i) {
//             pq.offer(new long[] {nums[i], i});
//         }
//         for (; k > 0 && pq.peek()[0] < m; --k) {
//             long[] p = pq.poll();
//             p[0] *= multiplier;
//             pq.offer(p);
//         }
//         final int mod = (int) 1e9 + 7;
//         for (int i = 0; i < n; ++i) {
//             long[] p = pq.poll();
//             long x = p[0];
//             int j = (int) p[1];
//             nums[j] = (int) ((x % mod) * qpow(multiplier, k / n + (i < k % n ? 1 : 0), mod) % mod);
//         }
//         return nums;
//     }
// 
//     private int qpow(long a, long n, long mod) {
//         long ans = 1 % mod;
//         for (; n > 0; n >>= 1) {
//             if ((n & 1) == 1) {
//                 ans = ans * a % mod;
//             }
//             a = a * a % mod;
//         }
//         return (int) 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 × log M + n × log k)

Space

O(n)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

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.