ShaysXIA/CodeInsightBench huggingface dataset:
Problem: 2063B - Subsequence Update
Language: java

Solution 1 (runtime: 593 ms):
import java.util.*;

public final class F1{
	public static void main(String[] args){
		Scanner sc = new Scanner(System.in);
		int t = sc.nextInt();
		while(t-- != 0){
			int n = sc.nextInt();
			int l = sc.nextInt();
			int r = sc.nextInt();
			int[] arr = new int[n];
			int[] left = new int[r];
			int[] right = new int[n-l+1];
			for(int i = 0; i < arr.length; i++){
				arr[i] = sc.nextInt();
			}
			for(int i = 0; i < left.length; i++){
				left[i] = arr[i];
			}
			for(int i = 0; i < right.length; i++){
				right[i] = arr[l-1+i];
			}
			Arrays.sort(left);
			Arrays.sort(right);
			long sum1 = 0;
			long sum2 = 0; 
			for(int i = 0 ; i <= r - l; i++){
				sum1 += (long)left[i];
				sum2 += (long)right[i];
			}
			System.out.println(Math.min(sum1, sum2));
		}
	}		
}

Solution 2 (runtime: 343 ms):
import java.util.*;

public class Solution {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();  // number of test cases
        while (t-- > 0) {
            // Read n, l, r
            int n = sc.nextInt();
            int l = sc.nextInt();  // 1-based index
            int r = sc.nextInt();  // 1-based index

            // Read array arr
            long[] arr = new long[n + 1];  // Using long[] for larger values, and n+1 to handle 1-based indexing
            for (int i = 1; i <= n; i++) {
                arr[i] = sc.nextLong();
            }

            // Step 1: Calculate the sum for the range [l, r]
            long initialSum = 0;
            for (int i = l; i <= r; i++) {
                initialSum += arr[i];
            }
            long min = initialSum;  // Store the initial sum as the minimum

            // Step 2: Find the smallest `r - l + 1` elements from [1, r]
            List<Long> prefix = new ArrayList<>();
            for (int i = 1; i <= r; i++) {
                prefix.add(arr[i]);
            }
            Collections.sort(prefix);
            long prefixSum = 0;
            // Sum the smallest `r - l + 1` elements from the sorted prefix
            for (int i = 0; i < r - l + 1; i++) {
                prefixSum += prefix.get(i);
            }
            min = Math.min(min, prefixSum);  // Update the minimum sum

            // Step 3: Find the smallest `r - l + 1` elements from [l, n]
            List<Long> suffix = new ArrayList<>();
            for (int i = l; i <= n; i++) {
                suffix.add(arr[i]);
            }
            Collections.sort(suffix);
            long suffixSum = 0;
            // Sum the smallest `r - l + 1` elements from the sorted suffix
            for (int i = 0; i < r - l + 1; i++) {
                suffixSum += suffix.get(i);
            }
            min = Math.min(min, suffixSum);  // Update the minimum sum

            // Output the result for the current test case
            System.out.println(min);
        }
    }
}