maximum sum of a subarray with at-least k elements.
// Returns maximum sum of a subarray with at-least // k elements. static int maxSumWithK(int a[], int n, int k) { // maxSum[i] is going to store maximum sum // till index i such that a[i] is part of the // sum. int maxSum[] = new int [n]; maxSum[0] = a[0]; // We use Kadane's algorithm to fill maxSum[] // Below code is taken from method 3 of // https://www.geeksforgeeks.org/largest-sum-contiguous-subarray/ int curr_max = a[0]; for (int i = 1; i < n; i++) { curr_max = Math.max(a[i], curr_max+a[i]); maxSum[i] = curr_max; } // Sum of first k elements int sum = 0; for (int i = 0; i < k; i++) sum += a[i]; // Use the concept of sliding window int result = sum; for (int i = k; i < n; i++) { // Compute sum of k elements ending // with a[i]. sum = sum + a[i] - a[i-k]; // Update result if required result = Math.max(result, sum); // Include maximum sum till [i-k] also // if it increases overall max. result = Math.max(result, sum + maxSum[i-k]); } return result; }