LeetCode #1524 — MEDIUM

Number of Sub-arrays With Odd Sum

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

Given an array of integers arr, return the number of subarrays with an odd sum.

Since the answer can be very large, return it modulo 10⁹ + 7.

Example 1:

Input: arr = [1,3,5]
Output: 4
Explanation: All subarrays are [[1],[1,3],[1,3,5],[3],[3,5],[5]]
All sub-arrays sum are [1,4,9,3,8,5].
Odd sums are [1,9,3,5] so the answer is 4.

Example 2:

Input: arr = [2,4,6]
Output: 0
Explanation: All subarrays are [[2],[2,4],[2,4,6],[4],[4,6],[6]]
All sub-arrays sum are [2,6,12,4,10,6].
All sub-arrays have even sum and the answer is 0.

Example 3:

Input: arr = [1,2,3,4,5,6,7]
Output: 16

Constraints:

1 <= arr.length <= 10⁵
1 <= arr[i] <= 100

Step 01

Brute Force Baseline

Problem summary: Given an array of integers arr, return the number of subarrays with an odd sum. Since the answer can be very large, return it modulo 109 + 7.

Baseline thinking

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

Pattern signal: Array · Math · Dynamic Programming

Example 1

[1,3,5]

Example 2

[2,4,6]

Example 3

[1,2,3,4,5,6,7]

Core Insight

What unlocks the optimal approach

Can we use the accumulative sum to keep track of all the odd-sum sub-arrays ?
if the current accu sum is odd, we care only about previous even accu sums and vice versa.

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

Upper-end input sizes

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 #1524: Number of Sub-arrays With Odd Sum
class Solution {
    public int numOfSubarrays(int[] arr) {
        final int mod = (int) 1e9 + 7;
        int[] cnt = {1, 0};
        int ans = 0, s = 0;
        for (int x : arr) {
            s += x;
            ans = (ans + cnt[s & 1 ^ 1]) % mod;
            ++cnt[s & 1];
        }
        return ans;
    }
}

// Accepted solution for LeetCode #1524: Number of Sub-arrays With Odd Sum
func numOfSubarrays(arr []int) (ans int) {
	const mod int = 1e9 + 7
	cnt := [2]int{1, 0}
	s := 0
	for _, x := range arr {
		s += x
		ans = (ans + cnt[s&1^1]) % mod
		cnt[s&1]++
	}
	return
}

# Accepted solution for LeetCode #1524: Number of Sub-arrays With Odd Sum
class Solution:
    def numOfSubarrays(self, arr: List[int]) -> int:
        mod = 10**9 + 7
        cnt = [1, 0]
        ans = s = 0
        for x in arr:
            s += x
            ans = (ans + cnt[s & 1 ^ 1]) % mod
            cnt[s & 1] += 1
        return ans

// Accepted solution for LeetCode #1524: Number of Sub-arrays With Odd Sum
impl Solution {
    pub fn num_of_subarrays(arr: Vec<i32>) -> i32 {
        const MOD: i32 = 1_000_000_007;
        let mut cnt = [1, 0];
        let mut ans = 0;
        let mut s = 0;
        for &x in arr.iter() {
            s += x;
            ans = (ans + cnt[((s & 1) ^ 1) as usize]) % MOD;
            cnt[(s & 1) as usize] += 1;
        }
        ans
    }
}

// Accepted solution for LeetCode #1524: Number of Sub-arrays With Odd Sum
function numOfSubarrays(arr: number[]): number {
    let ans = 0;
    let s = 0;
    const cnt: number[] = [1, 0];
    const mod = 1e9 + 7;
    for (const x of arr) {
        s += x;
        ans = (ans + cnt[(s & 1) ^ 1]) % mod;
        cnt[s & 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 × m)

Space

O(n × m)

Approach Breakdown

RECURSIVE

O(2ⁿ) time

O(n) space

Pure recursion explores every possible choice at each step. With two choices per state (take or skip), the decision tree has 2ⁿ leaves. The recursion stack uses O(n) space. Many subproblems are recomputed exponentially many times.

DYNAMIC PROGRAMMING

O(n × m) time

O(n × m) space

Each cell in the DP table is computed exactly once from previously solved subproblems. The table dimensions determine both time and space. Look for the state variables — each unique combination of state values is one cell. Often a rolling array can reduce space by one dimension.

Shortcut: Count your DP state dimensions → that’s your time. Can you drop one? That’s your space optimization.

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.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.

State misses one required dimension

Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.

Usually fails on: Correctness breaks on cases that differ only in hidden state.

Fix: Define state so each unique subproblem maps to one DP cell.