While preparing for coding tests or algorithm problem-solving, it is important to understand and learn approaches for various types of problems. Today, we will take an in-depth look at the problem of finding the sum of continuous elements.
Problem Description
We will examine the problem of calculating the sum of consecutive elements in a given integer array and finding the shortest continuous subarray that meets a specific criteria.
Problem Definition
Problem: Find the Sum of Consecutive Elements
Given an integer array nums
and an integer target
,
return the length of the shortest continuous subarray whose sum is greater than or equal to target
.
If there is no such subarray, return 0.
Input Example:
nums = [2,3,1,2,4,3]
target = 7
Output Example:
2 (length of subarray [4,3])
Problem Solving Approach
To solve this problem, two main approaches can be used: brute force (exhaustive search) and the two-pointer (sliding window) method. Here, we will use the two-pointer method for an efficient solution.
Two-Pointer Approach
The two-pointer approach involves exploring the array using two pointers (left and right) to find the subarray that satisfies the desired condition. The advantage of this method is its time complexity of O(n), making it efficient.
Step-by-Step Solution
-
Initialization: Initialize the two pointers. Set the left pointer
left
to 0 and the right pointerright
to 0. Also, initialize the current sumcurrentSum
to 0 and the minimum lengthminLength
to infinity. -
Condition Check: Traverse the array using the
right
pointer and addnums[right]
tocurrentSum
. Then check ifcurrentSum
is greater than or equal totarget
. -
Adjusting the Subarray: If
currentSum
is greater than or equal totarget
, update the minimum length and increase theleft
pointer while subtractingnums[left]
fromcurrentSum
. This helps find the shortest possible subarray. -
Termination Condition: Repeat this process until the
right
pointer reaches the end of the array.
Function Implementation
Now, let’s implement the above logic through C# code.
using System;
public class Solution {
public int MinSubArrayLen(int target, int[] nums) {
int left = 0;
int currentSum = 0;
int minLength = int.MaxValue;
for (int right = 0; right < nums.Length; right++) {
currentSum += nums[right];
while (currentSum >= target) {
minLength = Math.Min(minLength, right - left + 1);
currentSum -= nums[left];
left++;
}
}
return minLength == int.MaxValue ? 0 : minLength;
}
}
Code Explanation
The above C# code works as follows:
- Initial Variable Setup: Initialize
left
,currentSum
, andminLength
. - Array Traversal: Traverse the array using the
right
variable and add the current element tocurrentSum
. - Condition Check: If
currentSum
is greater than or equal totarget
, updateminLength
and increase theleft
pointer while subtractingnums[left]
fromcurrentSum
. - Return Result: Finally, if
minLength
has not been updated, return 0; otherwise, return the found minimum length.
Example Test Cases
Now, let’s write some example test cases to test this function.
public static void Main(string[] args) {
Solution sol = new Solution();
Console.WriteLine(sol.MinSubArrayLen(7, new int[] { 2, 3, 1, 2, 4, 3 })); // Output: 2
Console.WriteLine(sol.MinSubArrayLen(4, new int[] { 1, 4, 4 })); // Output: 1
Console.WriteLine(sol.MinSubArrayLen(11, new int[] { 1, 1, 1, 1, 1, 1 })); // Output: 0
Console.WriteLine(sol.MinSubArrayLen(8, new int[] { 2, 3, 1, 2, 4, 3 })); // Output: 2
}
Conclusion
In this lecture, we learned the problem-solving approach using two pointers through the problem of finding the sum of consecutive elements. It is important to accurately understand the nature of a problem and choose the appropriate approach to solve algorithmic problems. In practice, solving various problems helps gain experience and become familiar with different algorithms, so continuous practice is encouraged.
Try to enhance your algorithmic sense by encountering and solving various problems!