Find Longest Subarray with Given Sum using HashMap - Optimal Solution

Problem Statement❯

Given an array of integers (which may include positive numbers, negative numbers, or zeros) and a target sum k, your task is to find the length of the longest contiguous subarray that adds up exactly to k.

The subarray must be continuous, and if multiple such subarrays exist, return the length of the longest one.

If no such subarray exists, return 0.

Examples❯

Input Array	Target Sum (k)	Longest Subarray Length	Description
[1, 2, 3, -2, 1]	4	4	Subarray [1, 2, 3, -2] gives sum 4Visualization
[10, -10, 10, -10, 10]	0	4	Subarray [10, -10, 10, -10] gives sum 0Visualization
[5, 1, 1, 3, -2, 1]	4	5	Subarray [1, 1, 3, -2, 1] is the longest that gives sum 5Visualization
[1, 2, 3]	7	0	No subarray has sum 7Visualization
[0, 0, 0, 0]	0	4	Entire array sums to 0Visualization
[3, 4, -7, 1, 2]	0	3	Subarray [3, 4, -7] gives sum 0Visualization
[]	0	0	Empty array, so no subarray existsVisualization

Visualization Player

Solution❯

To find the longest subarray with a given sum k, especially when the array can have both positive and negative integers, we use a strategy based on prefix sums and a hash map.

Here’s the idea: As we traverse the array, we keep track of the cumulative sum (also called prefix sum) up to the current index. For each prefix sum, we store the first index where it occurred in a hash map.

Understanding the Intuition

Suppose at index i, the prefix sum is curr_sum. If there exists a prefix sum at an earlier index j such that:

curr_sum - prefix_sum[j] = k

then the subarray from index j+1 to i has a sum of k. By storing the first occurrence of each prefix sum in a map, we can look this up quickly and determine the length of such subarrays.

What Happens in Different Cases?

Exact match from the start: If the prefix sum itself becomes k, then the subarray from index 0 to i is valid. We track this using i + 1.
Zero sum cases: When the target sum is 0, the logic still works because the prefix sum might repeat, and the difference is 0.
All elements are zeros: If the entire array is zeros and the target is also zero, the whole array is the valid longest subarray.
No matching sum found: If we go through the array and never find a prefix sum that helps us build a valid subarray with sum = k, we return 0.
Empty array: If the array is empty, there’s no subarray at all, so the answer is 0.

Using this method ensures that we consider all possible subarrays without checking each one manually, and we do so efficiently with a time complexity of O(n).

This solution works great even with negative numbers, unlike sliding window techniques that require non-negative elements.

Algorithm Steps❯

Given an array arr of integers (positive, negative, or zero) and a target sum k.
Initialize prefix_sum = 0, max_len = 0, and an empty hash_map to store the first occurrence of each prefix sum.
Iterate through the array with index i:
→ Add arr[i] to prefix_sum.
→ If prefix_sum == k, update max_len = i + 1.
→ If prefix_sum - k exists in hash_map, calculate the length and update max_len accordingly.
→ If prefix_sum is not in hash_map, store prefix_sum with index i.
Return max_len after the loop ends.

Code

Python

JavaScript

Java

C++

def longest_subarray_with_sum_k(arr, k):
    prefix_sum = 0
    max_len = 0
    prefix_map = {}

    for i in range(len(arr)):
        prefix_sum += arr[i]

        if prefix_sum == k:
            max_len = i + 1

        if (prefix_sum - k) in prefix_map:
            max_len = max(max_len, i - prefix_map[prefix_sum - k])

        if prefix_sum not in prefix_map:
            prefix_map[prefix_sum] = i

    return max_len

# Sample Input
arr = [1, -1, 5, -2, 3]
k = 3
print("Length of Longest Subarray:", longest_subarray_with_sum_k(arr, k))

Time Complexity❯

Case	Time Complexity	Explanation
Best Case	O(n)	Each element is processed once, and HashMap lookups and insertions are O(1) on average. In the best case, the prefix sum quickly matches the target early in the array.
Average Case	O(n)	The algorithm traverses the entire array once, using a HashMap for fast prefix sum lookups and updates.
Worst Case	O(n)	Even in the worst case, the algorithm completes a single pass over the array, with constant-time operations per element.

Space Complexity❯

O(n)

Explanation: In the worst case, each prefix sum is unique and stored in the HashMap, resulting in O(n) additional space.

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