We have an array For the (Here, the given Return the answer to all queries. Your Example 1:<\/strong><\/p>\n\n\n\n Note:<\/strong><\/p>\n\n\n\n Time complexity: O(n + |Q|) Problem 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…<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[48],"tags":[20,222,179,62],"class_list":["post-4761","post","type-post","status-publish","format-standard","hentry","category-simulation","tag-array","tag-easy","tag-simulation","tag-sum","entry","simple"],"_links":{"self":[{"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/posts\/4761","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/comments?post=4761"}],"version-history":[{"count":5,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/posts\/4761\/revisions"}],"predecessor-version":[{"id":4767,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/posts\/4761\/revisions\/4767"}],"wp:attachment":[{"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/media?parent=4761"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/categories?post=4761"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/tags?post=4761"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}A<\/code> of integers, and an array queries<\/code> of queries.<\/p>\n\n\n\ni<\/code>-th query val = queries[i][0], index = queries[i][1]<\/code>, we add val to A[index]<\/code>. Then, the answer to the i<\/code>-th query is the sum of the even values of A<\/code>.<\/p>\n\n\n\nindex = queries[i][1]<\/code> is a 0-based index, and each query permanently modifies the array A<\/code>.)<\/em><\/p>\n\n\n\nanswer<\/code> array should have answer[i]<\/code> as the answer to the i<\/code>-th query.<\/p>\n\n\n\nInput: <\/strong>A = [1,2,3,4], queries = [[1,0],[-3,1],[-4,0],[2,3]]\nOutput: <\/strong>[8,6,2,4]\nExplanation: <\/strong>\nAt the beginning, the array is [1,2,3,4].\nAfter adding 1 to A[0], the array is [2,2,3,4], and the sum of even values is 2 + 2 + 4 = 8.\nAfter adding -3 to A[1], the array is [2,-1,3,4], and the sum of even values is 2 + 4 = 6.\nAfter adding -4 to A[0], the array is [-2,-1,3,4], and the sum of even values is -2 + 4 = 2.\nAfter adding 2 to A[3], the array is [-2,-1,3,6], and the sum of even values is -2 + 6 = 4.\n<\/pre>\n\n\n\n
1 <= A.length <= 10000<\/code><\/li>-10000 <= A[i] <= 10000<\/code><\/li>1 <= queries.length <= 10000<\/code><\/li>-10000 <= queries[i][0] <= 10000<\/code><\/li>0 <= queries[i][1] < A.length<\/code><\/li><\/ol>\n\n\n\nSolution: Simulation<\/h2>\n\n\n\n
Space complexity: O(n)<\/p>\n\n\n\nC++<\/h2>\n
\n\/\/ Author: Huahua, Running time: 100 ms \/ 6.5 MB\nclass Solution {\npublic:\n vectorPython3<\/h2>\n
\n# Author: Huahua, running time: 188 ms \/ 16.4MB\nclass Solution:\n def sumEvenAfterQueries(self, A: 'List[int]', queries: 'List[List[int]]') -> 'List[int]':\n s = sum(x for x in A if x % 2 == 0)\n ans = []\n for val, index in queries:\n if A[index] % 2 == 0: s -= A[index]\n A[index] += val\n if A[index] % 2 == 0: s += A[index]\n ans.append(s)\n return ans\n<\/pre>\n<\/div><\/div>\n","protected":false},"excerpt":{"rendered":"