The problem of calculating the sum of continuous subsequences in an array is a common topic in many coding tests. These problems are a good way to test the ability to manage control flow, manipulate arrays, and consider time complexity. In this tutorial, we will explore this topic in depth through the problem of ‘Calculating Continuous Sums’.<\/p>\n
Write a function to calculate the sum of Print the maximum sum of Explanation: The continuous 3 elements are (3, 4, 5) whose total sum is 12.<\/p>\n To solve this problem, there are two main approaches: calculating all combinations using simple iteration and an optimized approach using the sliding window technique.<\/p>\n While it is possible to calculate the sum of The sliding window technique works by maintaining the sum of Among the two methods above, we will choose the sliding window technique as it is more efficient and write the entire code.<\/p>\n In this tutorial, we examined how to calculate the sum of continuous subsequences in an array through the problem of 'Calculating Continuous Sums'. By using the sliding window technique, we can reduce the time complexity to O(n), allowing for a more efficient preparation for coding tests.<\/p>\n \n - Basic concepts for solving algorithm problems <\/body><\/p>\n","protected":false},"excerpt":{"rendered":"k<\/code> continuous elements from the given integer array arr<\/code>. If the length of arr<\/code> is n<\/code>, k<\/code> must be a value between 1<\/code> and n<\/code>. Here, the user should use an optimized method to calculate the continuous sum, and using simple loops is not recommended.<\/p>\nInput<\/h3>\n
\n
n<\/code> (1 \u2264 n \u2264 10^5<\/code>) and k<\/code> (1 \u2264 k \u2264 n<\/code>).<\/li>\narr<\/code> consisting of n<\/code> integers.<\/li>\n<\/ul>\nOutput<\/h3>\n
k<\/code> continuous elements.<\/p>\nExample<\/h3>\n
\n Input:\n 5 3\n 1 2 3 4 5\n\n Output:\n 12\n <\/pre>\n
Problem Solving Process<\/h2>\n
1. Simple Iteration Method<\/h3>\n
k<\/code> continuous elements using a simple iteration method, this approach has a time complexity of O(n*k) and becomes inefficient as the size of the array increases.<\/p>\n\n void simpleSum(int arr[], int n, int k) {\n int maxSum = 0;\n for (int i = 0; i <= n - k; i++) {\n int currentSum = 0;\n for (int j = 0; j < k; j++) {\n currentSum += arr[i + j];\n }\n maxSum = max(maxSum, currentSum);\n }\n cout << maxSum << endl;\n }\n <\/pre>\n2. Sliding Window Technique<\/h3>\n
k<\/code> elements by subtracting the previous value and adding the new value. This approach can reduce the time complexity to O(n).<\/p>\n\n void slidingWindowSum(int arr[], int n, int k) {\n int maxSum = 0, currentSum = 0;\n\n \/\/ Calculate the sum of the first k elements\n for (int i = 0; i < k; i++) {\n currentSum += arr[i];\n }\n maxSum = currentSum;\n\n \/\/ Calculate the sum through the sliding window\n for (int i = k; i < n; i++) {\n currentSum += arr[i] - arr[i - k];\n maxSum = max(maxSum, currentSum);\n }\n \n cout << maxSum << endl;\n }\n <\/pre>\nCode Implementation<\/h3>\n
\n #include
Conclusion<\/h2>\n
References<\/h3>\n
\n - Utilizing C++ STL
\n - Code optimization and testing patterns\n<\/p>\n