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Sum of Even Numbers After Queries - Array - Easy - LeetCode - MiniTV

Sum of Even Numbers After Queries - Array - Easy - LeetCode - मिनी टीवी

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 <= A.length <= 10000

-10000 <= A[i] <= 10000

1 <= queries.length <= 10000

-10000 <= queries[i][0] <= 10000

0 <= queries[i][1] < A.length

 

Solution:

using System;
using System.Collections.Generic;
using System.Text;

namespace LeetCode.AskGif.Easy.Array
{
    public class SumEvenAfterQueriesSoln
    {
        public int[] SumEvenAfterQueries(int[] A, int[][] queries)
        {
            var res = new int[A.Length];
            var sumEven = 0;
            int add = 0;
            for (int i = 0; i < A.Length; i++)
            {
                if(A[i] % 2 == 0)
                {
                    sumEven += A[i];
                }
            }

            for (int i = 0; i < A.Length; i++)
            {
                add = A[queries[i][1]] + queries[i][0];

                //delete old value in sumEven
                if (A[queries[i][1]] % 2 == 0)
                {
                    sumEven -= A[queries[i][1]];
                }

                //update new value in sumEven
                if(add % 2 == 0)
                {
                    sumEven += add;
                }

                //update array
                A[queries[i][1]] = add;

                res[i] = sumEven;
            }

            return res;
        }
    }
}

 

Time Complexity: O(n)

Space Complexity: O(1)

 

Unit Tests:

using LeetCode.AskGif.Easy.Array;
using Microsoft.VisualStudio.TestTools.UnitTesting;
using System;
using System.Collections.Generic;
using System.Text;

namespace CodingUnitTest.Easy.Array
{
    [TestClass]
    public class SumEvenAfterQueriesSolnTests
    {
        [TestMethod]
        public void SumEvenAfterQueriesSoln_First()
        {
            var A = new int[] { 1, 2, 3, 4 };
            var queries = new int[,] {
                {1, 0 },
                {-3, 1 },
                {-4, 0 },
                { 2, 3 }
            };
            var expected = new int[] { 8, 6, 2, 4 };

            var res = new SumEvenAfterQueriesSoln().SumEvenAfterQueries(A, ArrayMapper(queries));
            AreEqual(expected, res);
        }

        [TestMethod]
        public void SumEvenAfterQueriesSoln_Second()
        {
            var A = new int[] { 5, 5, 4 };
            var queries = new int[,] {                
                {0, 1 },
                {1, 0 },
                { 4, 1 }
            };
            var expected = new int[] { 4, 10, 10 };

            var res = new SumEvenAfterQueriesSoln().SumEvenAfterQueries(A, ArrayMapper(queries));
            AreEqual(expected, res);
        }

        private void AreEqual(int[] res, int[] output)
        {
            Assert.AreEqual(res.Length, output.Length);
            for (int i = 0; i < res.Length; i++)
            {
                Assert.AreEqual(res[i], output[i]);
            }
        }

        private int[][] ArrayMapper(int[,] matrix)
        {
            var arr = new int[matrix.GetLength(0)][];
            for (int i = 0; i < matrix.GetLength(0); i++)
            {
                arr[i] = new int[matrix.GetLength(1)];
                for (int j = 0; j < matrix.GetLength(1); j++)
                {
                    arr[i][j] = matrix[i, j];
                }
            }

            return arr;
        }
    }
}