Stickler thief and wants to loot money from a society of n houses placed in a line. He is a weird person and follows a rule while looting the houses and according to the rule Stickler thief will never loot two consecutive houses. At the same time, he wants to maximize the amount he loots. The thief knows which house has what amount of money but is unable to find the maximum amount he can end up with. He asks for your help to find the maximum money he can get if he strictly follows the rule. Each house has a[i] amount of money present in it.

**Input:**

The first line of input contains an integer T denoting the number of test cases. Each test case contains an integer n which denotes the number of elements in the array a[]. Next line contains space separated n elements in the array a[ ]

**Output:**

Print an integer which denotes the maximum amount he can take home.

**Constraints:**

1<=T<=200

1<=n<=1000

1<=a[i]<=10000

**Example:
Input:**

2

6

5 5 10 100 10 5

3

1 2 3

**Output:**

110

4

** solution:**

**Explanation:-** The money in each house will be array of integers as given in input example and we have to maximise this money without considering consecutive houses means we have to find maximum sum in array such that no two elements are adjacent.

**Simple Solution:-**

Loop for all elements in arr[] and maintain two sums sum1 and sum2l where sum1 = Max sum including the previous element and sum2 = Max sum excluding the previous element.

Max sum excluding the current element will be max(sum1, sum2) and max sum including the current element will be sum2 + current element (Note that only sum2 is considered because elements cannot be adjacent).

At the end of the loop return max of sum1 and sum2.

**Example:**

**arr[] = {5, 5, 10, 40, 50, 35}**

** sum1= 5 **

** sum2= 0**

** For i = 1 (current element is 5)**

** sum1= (sum2+ arr[i]) = 5**

** sum2= max(5, 0) = 5**

** For i = 2 (current element is 10)**

** sum1 = (sum2 + arr[i]) = 15**

** sum2 = max(5, 5) = 5**

** For i = 3 (current element is 40)**

** sum1 = (sum2+ arr[i]) = 45**

** sum2 = max(5, 15) = 15**

** For i = 4 (current element is 50)**

** sum1= (sum2 + arr[i]) = 65**

** sum2 = max(45, 15) = 45**

** For i = 5 (current element is 35)**

** sum1= (sum2 + arr[i]) = 80**

** sum2 = max(65, 45) = 65**

**And 35 is the last element. So, answer is max(sum1, sum2) = 80**

**Implementation:**

# include < bits / stdc++.h > using namespace std; int FindMaxSum(int arr[], int n) { int sum1 = arr[0]; int sum2 = 0; int result; for (int i = 1; i < n; i++){ result = (sum1 > sum2) ? sum1 : sum2; sum1 = sum2 + arr[i]; sum2 = result; } return ((sum1 > sum2) ? sum1 : sum2); } /* Driver program to test above function */ int main(){ int t; cin >> t; while (t–){ cin >> n; int arr[n]; for (int i = 0; i < n; i++) cin >> a[i]; cout << FindMaxSum(arr, n); } return 0; }

**DP Approach:**

**Explanation: **If there is only one house i.e. n=1 then simply return that if two houses are there that is n=2 then return max of the two otherwise repeat the similar method as explained in above method.

**implementation:**

# include < bits / stdc++.h > using namespace std; int funcMax(int a[], int n){ if (n == 1) return a[0]; if (n == 2) return max(a[0], a[1]); int * dp = (int * ) malloc(sizeof(int) * n); for (int i = 0; i < n; i++) dp[i] = 0; dp[0] = a[0]; dp[1] = max(a[0], a[1]); for (int i = 2; i < n; i++) dp[i] = max(dp[i - 2] + a[i], dp[i - 1]); int t = dp[n - 1]; free(dp); return t; } int main(){ int t; cin >> t; while (t–){ cin >> n; int arr[n]; for (int i = 0; i < n; i++) cin >> arr[i]; cout << funcMax(arr, n); } return 0; }