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1 | 1 | packagecom.fishercoder.solutions; |
2 | 2 |
|
3 | | -/** |
4 | | - * 985. Sum of Even Numbers After Queries |
5 | | - * |
6 | | - * We have an array A of integers, and an array queries of queries. |
7 | | - * 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. |
8 | | - * (Here, the given index = queries[i][1] is a 0-based index, and each query permanently modifies the array A.) |
9 | | - * Return the answer to all queries. Your answer array should have answer[i] as the answer to the i-th query. |
10 | | - * |
11 | | - * Example 1: |
12 | | - * |
13 | | - * Input: A = [1,2,3,4], queries = [[1,0],[-3,1],[-4,0],[2,3]] |
14 | | - * Output: [8,6,2,4] |
15 | | - * Explanation: |
16 | | - * At the beginning, the array is [1,2,3,4]. |
17 | | - * After adding 1 to A[0], the array is [2,2,3,4], and the sum of even values is 2 + 2 + 4 = 8. |
18 | | - * After adding -3 to A[1], the array is [2,-1,3,4], and the sum of even values is 2 + 4 = 6. |
19 | | - * After adding -4 to A[0], the array is [-2,-1,3,4], and the sum of even values is -2 + 4 = 2. |
20 | | - * After adding 2 to A[3], the array is [-2,-1,3,6], and the sum of even values is -2 + 6 = 4. |
21 | | - * |
22 | | - * Note: |
23 | | - * |
24 | | - * 1 <= A.length <= 10000 |
25 | | - * -10000 <= A[i] <= 10000 |
26 | | - * 1 <= queries.length <= 10000 |
27 | | - * -10000 <= queries[i][0] <= 10000 |
28 | | - * 0 <= queries[i][1] < A.length |
29 | | - */ |
30 | 3 | publicclass_985 { |
31 | | -publicstaticclassSolution1 { |
32 | | -publicint[]sumEvenAfterQueries(int[]A,int[][]queries) { |
33 | | -int[]result =newint[A.length]; |
34 | | -for (inti =0;i <A.length;i++) { |
35 | | -intcol =queries[i][1]; |
36 | | -A[col] =A[col] +queries[i][0]; |
37 | | -result[i] =computeEvenSum(A); |
38 | | - } |
39 | | -returnresult; |
40 | | - } |
| 4 | +publicstaticclassSolution1 { |
| 5 | +publicint[]sumEvenAfterQueries(int[]A,int[][]queries) { |
| 6 | +int[]result =newint[A.length]; |
| 7 | +for (inti =0;i <A.length;i++) { |
| 8 | +intcol =queries[i][1]; |
| 9 | +A[col] =A[col] +queries[i][0]; |
| 10 | +result[i] =computeEvenSum(A); |
| 11 | +} |
| 12 | +returnresult; |
| 13 | +} |
41 | 14 |
|
42 | | -privateintcomputeEvenSum(int[]A) { |
43 | | -intsum =0; |
44 | | -for (intnum :A) { |
45 | | -if (num %2 ==0) { |
46 | | -sum +=num; |
| 15 | +privateintcomputeEvenSum(int[]A) { |
| 16 | +intsum =0; |
| 17 | +for (intnum :A) { |
| 18 | +if (num %2 ==0) { |
| 19 | +sum +=num; |
| 20 | + } |
| 21 | + } |
| 22 | +returnsum; |
47 | 23 | } |
48 | | - } |
49 | | -returnsum; |
50 | 24 | } |
51 | | - } |
52 | 25 | } |