3637383940

package main

import (
	"fmt"
	"math"
)

// Function to find the longest subarray with a sum equal to k
func longestSubarray(nums []int, k int) int {
	n := len(nums)
	maxLength := 0

	// starting index
	for startIndex := 0; startIndex < n; startIndex++ {
		// ending index
		for endIndex := startIndex; endIndex < n; endIndex++ {
			// add all the elements of subarray = nums[startIndex...endIndex]
			currentSum := 0
			for i := startIndex; i <= endIndex; i++ {
				currentSum += nums[i]
			}

			if currentSum == k {
				maxLength = int(math.Max(float64(maxLength), float64(endIndex-startIndex+1)))
			}
		}
	}
	return maxLength
}

func main() {
	a := []int{-1, 1, 1}
	k := 1

	// Function call to get the result
	len := longestSubarray(a, k)

	fmt.Printf("The length of the longest subarray is: %d\n", len)
}

Complexity Analysis

Time Complexity: O(N³), where N is the size of the array. Since we are using three nested loops, each running approximately N times.

Space Complexity: O(1), as we are not using any extra space.

1d:T6025,

Intuition

In this approach, we use the concept of prefix sums to solve the problem of finding the longest subarray with a sum equal to 𝑘. The prefix sum of a subarray ending at index 𝑖 is simply the sum of all the elements of that subarray. By keeping track of these prefix sums and their occurrences, we can efficiently determine the longest subarray that meets the criteria.

Approach

First, we will declare a map to store the prefix sums and their corresponding indices.
We will run a loop from index 0 to n-1 (where n is the size of the array).
For each index i, add the current element a[i] to the prefix sum. If the prefix sum equals k, consider the length of the current subarray (i+1) and update the maximum length if needed. Calculate the prefix sum for the remaining subarray (prefix sum - k).
If this remaining sum exists in the map, calculate the length of the subarray ending at i and starting after the prefix sum, i.e., i - map[remaining sum], and update the maximum length if this length is greater. If the remaining sum does not exist in the map, add the current prefix sum and index to the map to ensure we store the earliest index where each prefix sum occurs.
The prefix sum of a subarray ending at index i is simply the sum of all the elements up to that index. By tracking these sums and their occurrences, we can efficiently determine the longest subarray that sums to k.

Observations

Assume the prefix sum of a subarray ending at index i is x. We need to find another subarray ending at i whose sum equals k. If such a subarray exists, the prefix sum of the remaining part will be x-k.

For a subarray ending at index i with the prefix sum x, if we remove the part with the prefix sum x-k, we are left with a subarray whose sum equals k. This is our target.

Instead of searching for subarrays with sum k, we will use a map to track the prefix sums at each index. The map will store each prefix sum and its corresponding index as a key-value pair. At index i, we will check the map for the prefix sum x-k. If it exists, we calculate the subarray length as i - preSumMap[x-k]. We update the maximum length if this subarray is longer.

We repeat this process for all indices from 0 to n-1 (where n is the size of the array).

Edge Case

Why do we need to check if the prefix sum already exists in the map? Let's un