codility test3

package com.lesson2;/** *  * You are given N counters, initially set to 0, and you have two possible * operations on them: increase(X) − counter X is increased by 1, max counter − * all counters are set to the maximum value of any counter. A non-empty * zero-indexed array A of M integers is given. This array represents * consecutive operations: if A[K] = X, such that 1 ≤ X ≤ N, then operation K is * increase(X), if A[K] = N + 1 then operation K is max counter. For example, * given integer N = 5 and array A such that: A[0] = 3 A[1] = 4 A[2] = 4 A[3] = * 6 A[4] = 1 A[5] = 4 A[6] = 4 the values of the counters after each * consecutive operation will be: (0, 0, 1, 0, 0) (0, 0, 1, 1, 0) (0, 0, 1, 2, * 0) (2, 2, 2, 2, 2) (3, 2, 2, 2, 2) (3, 2, 2, 3, 2) (3, 2, 2, 4, 2) The goal * is to calculate the value of every counter after all operations. Write a * function: class Solution { public int[] solution(int N, int[] A); } that, * given an integer N and a non-empty zero-indexed array A consisting of M * integers, returns a sequence of integers representing the values of the * counters. The sequence should be returned as: a structure Results (in C), or * a vector of integers (in C++), or a record Results (in Pascal), or an array * of integers (in any other programming language). For example, given: A[0] = 3 * A[1] = 4 A[2] = 4 A[3] = 6 A[4] = 1 A[5] = 4 A[6] = 4 the function should * return [3, 2, 2, 4, 2], as explained above. Assume that: N and M are integers * within the range [1..100,000]; each element of array A is an integer within * the range [1..N + 1]. Complexity: expected worst-case time complexity is * O(N+M); expected worst-case space complexity is O(N), beyond input storage * (not counting the storage required for input arguments). Elements of input * arrays can be modified. *  */public class MaxCounters {public int[] solution(int N, int[] A) {//correct,but not efficiencyint max = 0;int[] result = new int[N];for(int i = 0; i < A.length; i++){int ai = A[i];if(ai > N) {setResultToMax(result,max);} else {result[ai-1]++;if(result[ai-1] > max){max = result[ai-1];}}}return result;}private void setResultToMax(int[] result, int max) {for(int i = 0; i <result.length;i++){result[i] = max;}}public int[] solution1(int N, int[] A) {//correct,and efficiencyint max = 0;int preMax = 0;boolean hasSetToMax = false;int[] result = new int[N];for(int i = 0; i < A.length; i++){int ai = A[i];if(ai > N) {hasSetToMax = true;preMax = max;} else {if(hasSetToMax){if(result[ai-1] > preMax){result[ai-1]++;} else {result[ai-1] = preMax + 1;}} else {result[ai-1]++;}if(result[ai-1] > max){max = result[ai-1];}}}for(int i = 0; i <result.length;i++){if(result[i] < preMax){result[i] = preMax;}}return result;}public static void main(String[] args){MaxCounters counter = new MaxCounters();int[] A = {3,4,4,6,1,4,4};int[] result = counter.solution1(5, A);for(int i = 0; i <result.length;i++){System.out.println(result[i]);}}}