Review the fundamentals: Arrays guide and time complexities.

Count of Balanced Parity Subarrays — Problem Statement & Solution Guide

Q: How can we adapt the prefix sum technique to handle alternating sums where the parity of indices resets for each subarray?

We define a global prefix sum `S[i]` such that `S[i] = nums[0] - nums[1] + nums[2] - ... + (-1)^(i-1)nums[i-1]`. For a subarray starting at `j` and ending at `i-1`, the alternating sum is `(-1)^j * (S[i] - S[j])`. We then count pairs `(j, i)` such that `(-1)^j * (S[i] - S[j]) = k`. This splits into two cases based on the parity of `j`: if `j` is even, `S[i] - S[j] = k`; if `j` is odd, `S[i] - S[j] = -k`. We use two hash maps to store counts of prefix sums for even and odd starting indices separately.

ArraysMediumPrefix Sum / Hash Map

TimeO(N)

SpaceO(N)

Quick Answer & Algorithm Key Takeaway

Prefix Sum, Hash Map, Parity Tracking

TopicArrays

PatternPrefix Sum / Hash Map

TimeO(N)

SpaceO(N)

Problem Description

Given an integer array nums and an integer k, return the total number of continuous subarrays where the difference between the sum of elements at even relative indices and the sum of elements at odd relative indices is exactly k.

The relative index of an element in a subarray is its 0-based position within that subarray. For example, in the subarray [nums[i], nums[i+1], nums[i+2]], the elements at even relative indices are nums[i] and nums[i+2] (relative indices 0 and 2), and the element at the odd relative index is nums[i+1] (relative index 1).

DSA Pattern Breakdown

"Count of Balanced Parity Subarrays"

medium

WHY DOES IT MATTER?

This problem exemplifies the 'Prefix Sum with Parity' pattern, which is essential for solving problems involving alternating sums, weighted sums, or subarray properties that depend on index parity. It is a common variant of the standard prefix sum problem and appears frequently in interviews at top tech companies.

OPTIMIZATION CHALLENGE

The key optimization is recognizing that the alternating sum of a subarray can be expressed in terms of a global prefix sum with a sign factor that depends on the starting index's parity. This allows us to use hash maps to count valid pairs in O(N) time instead of O(N^2).

REAL-WORLD CONNECTION

In financial risk modeling, alternating sums can represent net exposure over time periods where gains and losses alternate. In signal processing, alternating sums are used in filter design where coefficients alternate in sign. In distributed systems, maintaining parity-aware state across nodes is crucial for consistent hashing and load balancing.

During the interview, clearly articulate the transformation from the subarray property to the prefix sum property. Emphasize the role of the sign factor (-1)^j and how it splits the problem into two independent counting problems. This demonstrates deep understanding of the pattern and its variations.

COMPLEXITY AT A GLANCE

⏱ Time:O(N)

💾 Space:O(N)

Core Theory — Why This Approach?

The problem requires counting subarrays where the alternating sum (sum of elements at even relative indices minus sum of elements at odd relative indices) equals a target value k. A naive approach would iterate over all O(N^2) subarrays and compute the alternating sum in O(N) time, resulting in O(N^3) complexity, which is infeasible for large inputs. The key insight is to transform the problem into a prefix sum problem by defining a modified prefix sum array P where P[i] represents the alternating sum of the subarray nums[0...i-1] with a specific sign convention that accounts for the relative index parity within any subarray starting at j.

Interview Questions on This Problem

Q1How can we adapt the prefix sum technique to handle alternating sums where the parity of indices resets for each subarray?

We define a global prefix sum S[i] such that S[i] = nums[0] - nums[1] + nums[2] - ... + (-1)^(i-1)nums[i-1]. For a subarray starting at j and ending at i-1, the alternating sum is (-1)^j * (S[i] - S[j]). We then count pairs (j, i) such that (-1)^j * (S[i] - S[j]) = k. This splits into two cases based on the parity of j: if j is even, S[i] - S[j] = k; if j is odd, S[i] - S[j] = -k. We use two hash maps to store counts of prefix sums for even and odd starting indices separately.

Q2In a fintech system, how would you handle this problem if the array represented transaction balances and `k` represented a net risk threshold?

The same algorithmic approach applies. The 'alternating sum' can be modeled as a weighted sum where weights alternate between +1 and -1. The prefix sum technique with hash maps remains optimal. In a distributed system, if the array is partitioned, we would need to merge prefix sum states from partitions, ensuring that the parity alignment is maintained across partition boundaries. This requires careful handling of the sign convention at the boundaries.

Q3What is the space complexity trade-off if we cannot use hash maps due to memory constraints in an embedded system?

Without hash maps, we would need to sort the prefix sums and use binary search or two pointers, but this only works if we are counting pairs with a fixed difference and the array is static. However, since the difference depends on the parity of the starting index, we would need to maintain two sorted lists of prefix sums (for even and odd starts). The time complexity would become O(N log N) due to sorting, and space complexity O(N) for the lists. This is less efficient than the O(N) time and O(N) space hash map approach but might be necessary in memory-constrained environments.

Examples

Example 1

Input

nums = [1, 2, 3, 1], k = 1

Output

Explanation: The 3 valid subarrays are: 1. [1] (index 0..0): Even-relative sum is 1, odd-relative sum is 0. Difference = 1. 2. [1] (index 3..3): Even-relative sum is 1, odd-relative sum is 0. Difference = 1. 3. [1, 2, 3, 1] (index 0..3): Even-relative sum is 1 + 3 = 4, odd-relative sum is 2 + 1 = 3. Difference = 1.

Example 2

Input

nums = [4, 2, 2], k = 2

Output

Explanation: The 3 valid subarrays are: 1. [4, 2] (index 0..1): Even-relative sum is 4, odd-relative sum is 2. Difference = 2. 2. [2] (index 1..1): Even-relative sum is 2, odd-relative sum is 0. Difference = 2. 3. [2] (index 2..2): Even-relative sum is 2, odd-relative sum is 0. Difference = 2.

Constraints

1 <= nums.length <= 10^5
-10^4 <= nums[i] <= 10^4
-10^9 <= k <= 10^9

Optimal Approach & Strategy

Compute a global prefix sum array where each element is the alternating sum of the array up to that point. Use two hash maps to store the counts of prefix sums for even and odd starting indices. For each index, check how many previous prefix sums satisfy the conditions prefix[i] - prefix[j] = k (for even j) or prefix[i] - prefix[j] = -k (for odd j). This runs in O(N) time and O(N) space.

Brute Force Approach

Iterate over all possible starting indices i and ending indices j, and for each subarray, compute the alternating sum by iterating through the subarray elements. This results in O(N^3) time complexity, which is too slow for large inputs.

Code Solutions

Language:

JavaScript Solution

Time: O(N)

function countOfBalancedParitySubarrays(nums, k) {
    let count = 0, currentSum = 0;
    let map = new Map([[0, 1]]);
    for (let i = 0; i < nums.length; i++) {
        let val = i % 2 === 0 ? nums[i] : -nums[i];
        currentSum += val;
        if (map.has(currentSum - k)) count += map.get(currentSum - k);
        map.set(currentSum, (map.get(currentSum) || 0) + 1);
    }
    return count;
}

#include <vector>
#include <unordered_map>

class Solution {
public:
    long long countOfBalancedParitySubarrays(std::vector<int>& nums, int k) {
        long long count = 0;
        long long currentSum = 0;
        std::unordered_map<long long, int> even_map;
        std::unordered_map<long long, int> odd_map;
        
        even_map[0] = 1;
        
        for (int j = 1; j <= nums.size(); ++j) {
            int sign = ((j - 1) % 2 == 0) ? 1 : -1;
            currentSum += (long long)sign * nums[j - 1];
            
            auto it_even = even_map.find(currentSum - k);
            if (it_even != even_map.end()) {
                count += it_even->second;
            }
            
            auto it_odd = odd_map.find(currentSum + k);
            if (it_odd != odd_map.end()) {
                count += it_odd->second;
            }
            
            if (j % 2 == 0) {
                even_map[currentSum]++;
            } else {
                odd_map[currentSum]++;
            }
        }
        
        return count;
    }
};

import java.util.HashMap;
import java.util.Map;

class Solution {
    public long countOfBalancedParitySubarrays(int[] nums, int k) {
        long count = 0;
        long currentPrefixSum = 0;
        // We track the parity of the starting index relative to the global array.
        // If a subarray starts at index 'i', the relative index 0 corresponds to 
        // absolute index 'i'. If 'i' is even, absolute even indices are relative even.
        // If 'i' is odd, absolute even indices are relative odd.
        Map<Long, Integer> evenStartMap = new HashMap<>();
        Map<Long, Integer> oddStartMap = new HashMap<>();
        
        evenStartMap.put(0L, 1);
        
        for (int i = 0; i < nums.length; i++) {
            currentPrefixSum += (i % 2 == 0) ? nums[i] : -nums[i];
            
            if (i % 2 == 0) {
                // currentPrefixSum - prev = k => prev = currentPrefixSum - k
                count += evenStartMap.getOrDefault(currentPrefixSum - k, 0);
                evenStartMap.put(currentPrefixSum, evenStartMap.getOrDefault(currentPrefixSum, 0) + 1);
            } else {
                // If starting at an odd index, signs are flipped relative to the global sum
                // -(currentPrefixSum - prev) = k => prev = currentPrefixSum + k
                count += oddStartMap.getOrDefault(currentPrefixSum + k, 0);
                oddStartMap.put(currentPrefixSum, oddStartMap.getOrDefault(currentPrefixSum, 0) + 1);
            }
        }
        return count;
    }
}

def count_of_balanced_parity_subarrays(nums, k):
    from collections import defaultdict
    even_map = defaultdict(int)
    odd_map = defaultdict(int)
    
    current_sum = 0
    count = 0
    
    for x, val in enumerate(nums):
        if x % 2 == 0:
            even_map[current_sum] += 1
            current_sum += val
        else:
            odd_map[current_sum] += 1
            current_sum -= val
            
        count += even_map[current_sum - k]
        count += odd_map[current_sum + k]
        
    return count

function countOfBalancedParitySubarrays(nums, k) {
    let count = 0, currentSum = 0;
    let map = new Map([[0, 1]]);
    for (let i = 0; i < nums.length; i++) {
        let val = i % 2 === 0 ? nums[i] : -nums[i];
        currentSum += val;
        if (map.has(currentSum - k)) count += map.get(currentSum - k);
        map.set(currentSum, (map.get(currentSum) || 0) + 1);
    }
    return count;
}

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