#985 - Sum of Even Numbers After Queries

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Problem 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.

Solution

/**
 * @param {number[]} A
 * @param {number[][]} queries
 * @return {number[]}
 */
var sumEvenAfterQueries = function(A, queries) {
  return queries.reduce((sums, query) => {
    A[query[1]] += query[0];
    return [...sums, A.reduce((sum, n) => sum + (n % 2 === 0 ? n : 0), 0)];
  }, []);
};

// Significantly faster, less elegant approach:
/**
 * @param {number[]} A
 * @param {number[][]} queries
 * @return {number[]}
 */
var sumEvenAfterQueries = function(A, queries) {
  return queries.reduce((sums, query) => {
    const [v, i] = query;
    const prev = A[i];
    const lastSum = sums.length ? sums[sums.length - 1] : A.reduce((sum, n) => sum + (n % 2 === 0 ? n : 0), 0);
    A[i] += v;
    sums.push(lastSum - (prev % 2 === 0 ? prev : 0) + (A[i] % 2 === 0 ? A[i] : 0));
    return sums;
  }, []);
};