(Easy) Sum of Even Numbers After Queries - LeetCode

Description:

We have an array A of integers, and an array queries of queries.

For the i-th query val = queries[i][0], index = queries[i][1], we add val to A[index].  Then, the answer to the i-th query is the sum of the even values of A.

(Here, the given index = queries[i][1] is a 0-based index, and each query permanently modifies the array A.)

Return the answer to all queries.  Your answer array should have answer[i] as the answer to the i-th query.

 

Example 1:

Input: A = [1,2,3,4], queries = [[1,0],[-3,1],[-4,0],[2,3]]
Output: [8,6,2,4]
Explanation: 
At the beginning, the array is [1,2,3,4].
After adding 1 to A[0], the array is [2,2,3,4], and the sum of even values is 2 + 2 + 4 = 8.
After adding -3 to A[1], the array is [2,-1,3,4], and the sum of even values is 2 + 4 = 6.
After adding -4 to A[0], the array is [-2,-1,3,4], and the sum of even values is -2 + 4 = 2.
After adding 2 to A[3], the array is [-2,-1,3,6], and the sum of even values is -2 + 6 = 4.

 

Note:

1. 1 <= A.length <= 10000
2. -10000 <= A[i] <= 10000
3. 1 <= queries.length <= 10000
4. -10000 <= queries[i][0] <= 10000
5. 0 <= queries[i][1] < A.length

Accepted

26,854

Submissions

42,932

Solution:

 

 

Attempt 1. failed due to Time out exception:

class Solution {
    public int[] sumEvenAfterQueries(int[] A, int[][] queries) {
        
        if(A==null||A.length==0){
            return null; 
        }
        
        if(queries==null||queries.length==0){
            return A;
        }
        
        int [] B = new int[A.length];
        
        for(int i = 0; i<queries.length; i++){
            
            int val = queries[i][0];
            int pos = queries[i][1];
            
            A[pos] = A[pos]+val; 
            
            B[i] = Sum(A);
            
        }
        
        return B; 
        
    }
    
    public int Sum (int[] arr){
        int sum = 0; 
        for(int i = 0; i<arr.length; i++){
            
            if(Math.abs(arr[i])%2==0){
                sum = sum +arr[i]; 
                
            }
        }
        
        return sum; 
    }
}

 

Successful Attempt. Accepted. 

class Solution {
    public int[] sumEvenAfterQueries(int[] A, int[][] queries) {
        
        if(A==null||A.length==0){
            return null; 
        }
        
        if(queries==null||queries.length==0){
            return A;
        }
        
        int [] B = new int[A.length];
         int sum = 0; 
        for(int i = 0; i<A.length; i++){
            
            if(Math.abs(A[i])%2==0){
                sum = sum +A[i]; 
                
            }
        }
        System.out.println(sum);
        
        for(int i = 0; i<queries.length; i++){
            
            int val = queries[i][0];
            int pos = queries[i][1];
        
           
            
            if(Math.abs(A[pos]+val)%2==0){
                
                if(Math.abs(A[pos])%2!=0){
                    B[i] = sum+A[pos]+val;   
                     sum = sum + A[pos]+ val;
                   
                 // System.out.println(i+ " "+sum + "Check 1 "+ val+" "+pos+" "+A[pos]+" "+ B[i]);
                } 
                
                else{
                    
                    B[i] = sum+val;
                    sum = sum + val;
                     //System.out.println("Check 2 "+ "val = " +val+" "+" pos = "+pos+" "+A[pos]+" "+ B[i]);
                }
                
                
                 A[pos] = A[pos]+val;
        
            }
            
            else{
                if(Math.abs(A[pos])%2==0){
                      B[i] = sum - A[pos];
                
                        sum = sum - A[pos];
                        A[pos] = A[pos]+val;
                }
                
                else{
                    
                    B[i] = sum; 
                    
                    A[pos] = A[pos]+val;
                }
              
            }
            
        }
        
        return B; 
        
    }
    
    
}