ICSE Computer Applications previous year question papers, solved by the pattern they follow.

import java.util.Scanner;
class SwapEnds {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        System.out.print("Enter a sentence: ");
        String s = sc.nextLine().trim() + " ";
        String word = "", result = "";
        int vowelStart = 0;
        for (int i = 0; i < s.length(); i++) {
            char ch = s.charAt(i);
            if (ch != ' ') {
                word = word + ch;
            } else if (word.length() > 0) {
                char f = Character.toUpperCase(word.charAt(0));
                if ("AEIOU".indexOf(f) >= 0)
                    vowelStart++;
                int L = word.length();
                if (L > 1)
                    word = word.charAt(L - 1) + word.substring(1, L - 1) + word.charAt(0);
                result = result + word + " ";
                word = "";
            }
        }
        System.out.println(result.trim());
        System.out.println("Words starting with a vowel: " + vowelStart);
    }
}

eht mxae si rean
Words starting with a vowel: 2

Adding one space to the end means the last word is handled by the same branch as the others. The check word.length() > 0 skips double spaces, and L > 1 leaves a one-letter word such as "a" alone, where substring(1, 0) would crash.

import java.util.Scanner;
class Symmetric {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int[][] a = new int[3][3];
        System.out.println("Enter 9 elements:");
        for (int i = 0; i < 3; i++)
            for (int j = 0; j < 3; j++)
                a[i][j] = sc.nextInt();
        boolean sym = true;
        int boundary = 0;
        for (int i = 0; i < 3; i++)
            for (int j = 0; j < 3; j++) {
                if (a[i][j] != a[j][i])
                    sym = false;
                if (i == 0 || i == 2 || j == 0 || j == 2)
                    boundary += a[i][j];
            }
        System.out.println(sym ? "Symmetric matrix" : "Not a symmetric matrix");
        System.out.println("Sum of boundary elements = " + boundary);
    }
}

// input 1 4 7 / 4 2 5 / 7 5 3
Symmetric matrix
Sum of boundary elements = 36

A boundary cell is in the first or last row or column; in a 3 by 3 matrix only the centre is left out. Input 1 to 9 in order prints Not a symmetric matrix and 40.

import java.util.Scanner;
class CitySearch {
    public static void main(String[] args) {
        String[] c = {"Pune", "Agra", "Surat", "Kochi", "Delhi", "Bhopal"};
        int n = c.length;
        for (int i = 0; i < n - 1; i++)
            for (int j = 0; j < n - 1 - i; j++)
                if (c[j].compareToIgnoreCase(c[j + 1]) > 0) {
                    String t = c[j]; c[j] = c[j + 1]; c[j + 1] = t;
                }
        for (int i = 0; i < n; i++)
            System.out.print(c[i] + " ");
        System.out.println();
        Scanner sc = new Scanner(System.in);
        System.out.print("City to search: ");
        String key = sc.next();
        int low = 0, high = n - 1, pos = -1;
        while (low <= high) {
            int mid = (low + high) / 2;
            int r = key.compareToIgnoreCase(c[mid]);
            if (r == 0) { pos = mid; break; }
            else if (r < 0) high = mid - 1;
            else low = mid + 1;
        }
        if (pos >= 0)
            System.out.println(key + " found at index " + pos);
        else
            System.out.println(key + " not found");
    }
}

Agra Bhopal Delhi Kochi Pune Surat
surat found at index 5

Agra Bhopal Delhi Kochi Pune Surat
Goa not found

Binary search only works on sorted data, which is why the sort comes first. Strings are compared with compareToIgnoreCase, never with < or ==, and the sign of the result says which half to keep.