Evening Harmony Partition Count

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In Evening Harmony Partition Count you are given an array of note volumes and a limit You need to count how many ways the array can be split into contiguous segments such that every segment sum is at most the given limit The partition must cover the full array in order and each cut creates a new segment boundary If any required segment would exceed the limit that partition path is invalid This is a dynamic-programming counting problem A clean setup is dp i number of valid ways to partition the prefix ending before index i From each start point expand a running sum forward until it exceeds the limit every valid end contributes transitions to the next state Since counts can become huge always reduce with modulo 1 000 000 007 For an empty array there is exactly one valid partition choose no segments The judge includes scenarios with dense valid cuts strict limits causing zero ways and small arrays where all segment choices are valid Output must be a single integer count only This task is excellent for practicing prefix-style DP transitions window expansion for sum constraints and careful handling of edge cases like empty input and values that individually exceed the segment limit Use running sums to keep transition checks fast on long note arrays When the task involves connectivity or route cost build adjacency carefully and guard against revisiting stale states Use a visited or best-distance

Examples

Example 1

Input

notes = [5], limit = 3

Output

Explanation

The single note exceeds the volume limit, so no partition is possible.

Example 2

Input

notes = [2,2,2,2], limit = 4

Output

Explanation

Each segment may contain one or two notes, yielding five partitions.

Example 3

Input

notes = [1,2,1], limit = 3

Output

Explanation

Valid segmentations are [1|2|1], [1|2,1], and [1,2|1].

Algorithm Flow

Recommendation Algorithm Flow for Evening Harmony Partition Count - Budibadu

Best Answers

java

import java.util.*;
class Solution {
    public int count_partitions(int[] nums, int k) {
        long total = 0;
        for (int x : nums) total += x;
        if (total % k != 0) return 0;
        long target = total / k;
        int[] ways = new int[k];
        ways[0] = 1;
        long current = 0;
        for (int i = 0; i < nums.length - 1; i++) {
            current += nums[i];
            for (int j = k - 1; j > 0; j--) {
                if (current == j * target) {
                    ways[j] += ways[j-1];
                }
            }
        }
        return ways[k-1];
    }
}

javascript

function count_partitions(nums, k) {
    const total = nums.reduce((a, b) => a + b, 0);
    if (total % k !== 0) return 0;
    const target = total / k;
    const ways = new Array(k).fill(0);
    ways[0] = 1;
    let current = 0;
    for (let i = 0; i < nums.length - 1; i++) {
        current += nums[i];
        for (let j = k - 1; j > 0; j--) {
            if (current === j * target) {
                ways[j] += ways[j-1];
            }
        }
    }
    return ways[k-1];
}

php

function count_partitions($nums, $k) {
    $total = array_sum($nums);
    if ($total % $k !== 0) return 0;
    $target = $total / $k;
    $ways = array_fill(0, $k, 0);
    $ways[0] = 1;
    $current = 0;
    for ($i = 0; $i < count($nums) - 1; $i++) {
        $current += $nums[$i];
        for ($j = $k - 1; $j > 0; $j--) {
            if ($current == $j * $target) {
                $ways[$j] += $ways[$j-1];
            }
        }
    }
    return $ways[$k-1];
}

python

def count_partitions(nums, k):
    n = len(nums)
    total = sum(nums)
    if total % k != 0: return 0
    target = total // k
    if target == 0:
        zeros = nums.count(0)
        if zeros < k: return 0
        # Ways to pick k-1 zeros to split among total zeros
        import math
        return math.comb(zeros - 1, k - 1)
    
    count = 0
    ways = [0] * (k + 1)
    ways[0] = 1
    current = 0
    for i in range(n - 1):
        current += nums[i]
        for j in range(k - 1, 0, -1):
            if current == j * target:
                ways[j] += ways[j-1]
    return ways[k-1]

rust

pub fn count_partitions(nums: Vec<i32>, k: i32) -> i32 {
    let total: i64 = nums.iter().map(|&x| x as i64).sum();
    let k = k as usize;
    if total % (k as i64) != 0 { return 0; }
    let target = total / (k as i64);
    let mut ways = vec![0; k];
    ways[0] = 1;
    let mut current = 0i64;
    for i in 0..nums.len().saturating_sub(1) {
        current += nums[i] as i64;
        for j in (1..k).rev() {
            if current == (j as i64) * target {
                ways[j] += ways[j-1];
            }
        }
    }
    ways[k-1]
}

typescript

function count_partitions(nums: number[], k: number): number {
    const total = nums.reduce((a, b) => a + b, 0);
    if (total % k !== 0) return 0;
    const target = total / k;
    const ways = new Array(k).fill(0);
    ways[0] = 1;
    let current = 0;
    for (let i = 0; i < nums.length - 1; i++) {
        current += nums[i];
        for (let j = k - 1; j > 0; j--) {
            if (current === j * target) {
                ways[j] += ways[j-1];
            }
        }
    }
    return ways[k-1];
}

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Evening Harmony Partition Count

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