clang formatted

Author: Guhapriya01

src/main/java/com/thealgorithms/dynamicprogramming/MaximumSumOfNonAdjacentElements.java

@@ -7,79 +7,79 @@
  * https://takeuforward.org/data-structure/maximum-sum-of-non-adjacent-elements-dp-5/
  */
 final class MaximumSumOfNonAdjacentElements {
-	private MaximumSumOfNonAdjacentElements() {
-	}
-
-	/**
-	 * Approach 1: Uses a dynamic programming array to store the maximum sum at each
-	 * index. Time Complexity: O(n) - where n is the length of the input array.
-	 * Space Complexity: O(n) - due to the additional dp array.
-	 * @param arr The input array of integers.
-	 * @return The maximum sum of non-adjacent elements.
-	 */
-	public static int getMaxSumApproach1(int[] arr) {
-		int n = arr.length;
-		int[] dp = new int[n];
-
-		// Base case: Maximum sum if only one element is present.
-		dp[0] = arr[0];
-
-		for (int ind = 1; ind < n; ind++) {
-
-			// Case 1: Do not take the current element, carry forward the previous max sum.
-			int notTake = dp[ind - 1];
-
-			// Case 2: Take the current element, add it to the max sum up to two indices
-			// before.
-			int take = arr[ind];
-			if (ind > 1) {
-				take += dp[ind - 2];
-			}
-
-			// Store the maximum of both choices in the dp array.
-			dp[ind] = Math.max(take, notTake);
-		}
-
-		return dp[n - 1];
-	}
-
-	/**
-	 * Approach 2: Optimized space complexity approach using two variables instead
-	 * of an array. Time Complexity: O(n) - where n is the length of the input
-	 * array. Space Complexity: O(1) - as it only uses constant space for two
-	 * variables.
-	 * @param arr The input array of integers.
-	 * @return The maximum sum of non-adjacent elements.
-	 */
-	public static int getMaxSumApproach2(int[] arr) {
-		int n = arr.length;
-
-		// Two variables to keep track of previous two results:
-		// prev1 = max sum up to the last element (n-1)
-		// prev2 = max sum up to the element before last (n-2)
-
-		int prev1 = arr[0]; // Base case: Maximum sum for the first element.
-		int prev2 = 0;
-
-		for (int ind = 1; ind < n; ind++) {
-			// Case 1: Do not take the current element, keep the last max sum.
-			int notTake = prev1;
-
-			// Case 2: Take the current element and add it to the result from two steps
-			// back.
-			int take = arr[ind];
-			if (ind > 1) {
-				take += prev2;
-			}
-
-			// Calculate the current maximum sum and update previous values.
-			int current = Math.max(take, notTake);
-
-			// Shift prev1 and prev2 for the next iteration.
-			prev2 = prev1;
-			prev1 = current;
-		}
-
-		return prev1;
-	}
-}
+  private MaximumSumOfNonAdjacentElements() {}
+
+  /**
+   * Approach 1: Uses a dynamic programming array to store the maximum sum at
+   * each index. Time Complexity: O(n) - where n is the length of the input
+   * array. Space Complexity: O(n) - due to the additional dp array.
+   * @param arr The input array of integers.
+   * @return The maximum sum of non-adjacent elements.
+   */
+  public static int getMaxSumApproach1(int[] arr) {
+    int n = arr.length;
+    int[] dp = new int[n];
+
+    // Base case: Maximum sum if only one element is present.
+    dp[0] = arr[0];
+
+    for (int ind = 1; ind < n; ind++) {
+
+      // Case 1: Do not take the current element, carry forward the previous max
+      // sum.
+      int notTake = dp[ind - 1];
+
+      // Case 2: Take the current element, add it to the max sum up to two
+      // indices before.
+      int take = arr[ind];
+      if (ind > 1) {
+        take += dp[ind - 2];
+      }
+
+      // Store the maximum of both choices in the dp array.
+      dp[ind] = Math.max(take, notTake);
+    }
+
+    return dp[n - 1];
+  }
+
+  /**
+   * Approach 2: Optimized space complexity approach using two variables instead
+   * of an array. Time Complexity: O(n) - where n is the length of the input
+   * array. Space Complexity: O(1) - as it only uses constant space for two
+   * variables.
+   * @param arr The input array of integers.
+   * @return The maximum sum of non-adjacent elements.
+   */
+  public static int getMaxSumApproach2(int[] arr) {
+    int n = arr.length;
+
+    // Two variables to keep track of previous two results:
+    // prev1 = max sum up to the last element (n-1)
+    // prev2 = max sum up to the element before last (n-2)
+
+    int prev1 = arr[0]; // Base case: Maximum sum for the first element.
+    int prev2 = 0;
+
+    for (int ind = 1; ind < n; ind++) {
+      // Case 1: Do not take the current element, keep the last max sum.
+      int notTake = prev1;
+
+      // Case 2: Take the current element and add it to the result from two
+      // steps back.
+      int take = arr[ind];
+      if (ind > 1) {
+        take += prev2;
+      }
+
+      // Calculate the current maximum sum and update previous values.
+      int current = Math.max(take, notTake);
+
+      // Shift prev1 and prev2 for the next iteration.
+      prev2 = prev1;
+      prev1 = current;
+    }
+
+    return prev1;
+  }
+}