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Rearrange Array in Maximum-Minimum form

Given a sorted array of positive integers, rearrange the array alternately i.e first element should be the maximum value, second minimum value, third-second max, fourth-second min and so on. Naive Solution: Here we will use an auxiliary array to store the elements in correct order. For even indices of the array  Large numbers from the given array are added. For odd indices small numbers are added. Code: import java.util.*; public class Test{        public static void main(String[] args) {          Scanner in = new Scanner(System.in);     int[] a = {1,2,3,4,5,6,7,8,9};     int n = a.length;     int[] temp = new int[n];     int small = 0;     int large = n-1;     for(int i=0; i<n; i++){         if(i%2 == 0){             temp[i] = a[large--];         }  ...
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Reverse words in a given string

Given a String of length N reverse the whole string without reversing the individual words in it. Words are separated by dots. Code: package sap; public class ReverseWords {         public static void main(String[] args){                 String word = "I am done";         StringBuilder result = new StringBuilder("");                 String[] arr = word.split("\\s");                 for(int i=arr.length-1; i>=0; i--){                         result.append(arr[i]);             result.append(" ");                     }   ...
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Second Largest Element

Given an array of integers, our task is to write a program that efficiently finds the second largest element present in the array. Example: Input : arr[] = {12, 35, 1, 10, 34, 1} Output : The second largest element is 34. Code: package sap; public class SecondLargest {         public static void main(String[] args){                 int[] a = {89, 24, 75, 11, 23};         int max = Integer.MIN_VALUE, secondMax = Integer.MIN_VALUE;                 for(int i=0; i<a.length; i++){                         if(a[i] > max){                                 secondMax = max;     ...
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