Arrays ICSE Class 10 Computer Applications: Sorting and Searching

What Arrays is about

Weightage: CISCE publishes no chapter weightage. Section A (40 marks) is compulsory: 1-mark MCQs and 2-mark short answers including output and error questions; in Section B you answer any four of six 15-mark programming questions written in Java.Source: CISCE ICSE Computer Applications (661) specimen paper structure for 2027

Key concepts

Single dimensional arrays

An array is a collection of variables of the same data type stored under one name in consecutive memory locations; each element is reached by an index (subscript) from 0 to length − 1. It is a composite (non-primitive) type.

Declaration and creation: int a[] = new int[5]; (5 ints, all 0). Or int[] a = new int[5];
Initialisation: int a[] = {4, 9, 2, 7, 9, 1}; String s[] = {"Blue", "Red"};
Access: a[0] is 4; a[a.length - 1] is the last element.
length: a.length gives the number of elements (no brackets; for a String it is s.length() with brackets).

Input with a loop:
for (int i = 0; i < a.length; i++)
a[i] = sc.nextInt();

Memory: elements × size of type: double d[] = new double[4] uses 4 × 8 = 32 bytes.

Using an index outside 0 to length − 1 (a[6] for a 6-element array) causes a runtime error (ArrayIndexOutOfBoundsException).

Worked example

(2 marks) int a[] = {4, 9, 2, 7, 9, 1}; (a) What is a[1] + a[4]? (b) What is the index of the smallest element?

Try it first, then show the solutionSolution

1. a[1] = 9 and a[4] = 9; sum 18 - 1 mark.
2. Smallest element 1 is at index 5 - 1 mark.

Answer: (a) 18 (b) 5

Selection sort

Selection sort: in each pass, find the smallest element in the unsorted part and swap it with the first element of that part. After pass i, the first i + 1 elements are in their final places.

for (int i = 0; i < n - 1; i++)
{
int min = i;
for (int j = i + 1; j < n; j++)
if (a[j] < a[min])
min = j;
int t = a[i];
a[i] = a[min];
a[min] = t;
}

Trace for {64, 25, 12, 22, 11}:
Pass 1: smallest 11 (index 4) swapped with 64 → 11 25 12 22 64
Pass 2: smallest of 25 12 22 64 is 12 → 11 12 25 22 64
Pass 3: smallest of 25 22 64 is 22 → 11 12 22 25 64
Pass 4: smallest of 25 64 is 25 (no change) → 11 12 22 25 64

For descending order, look for the largest (a[j] > a[max]).

Worked example

(2 marks) Show the array {9, 4, 7, 1} after the first two passes of selection sort (ascending).

Try it first, then show the solutionSolution

1. Pass 1: smallest 1 swapped with 9 → 1 4 7 9 - 1 mark.
2. Pass 2: smallest of 4 7 9 is 4, already first → 1 4 7 9 - 1 mark.

Answer: After pass 1: 1 4 7 9; after pass 2: 1 4 7 9.

Bubble sort

Bubble sort: compare each pair of adjacent elements and swap them if they are in the wrong order; after each pass the largest remaining element 'bubbles' to the end.

for (int i = 0; i < n - 1; i++)
for (int j = 0; j < n - 1 - i; j++)
if (a[j] > a[j + 1])
{
int t = a[j];
a[j] = a[j + 1];
a[j + 1] = t;
}

Trace for {5, 1, 4, 2, 8}:
Pass 1: (5,1) swap → 1 5 4 2 8; (5,4) swap → 1 4 5 2 8; (5,2) swap → 1 4 2 5 8; (5,8) no swap.
Pass 2: (1,4) no; (4,2) swap → 1 2 4 5 8; (4,5) no.
Pass 3 and 4: no swaps. Result: 1 2 4 5 8.

Selection vs bubble: selection makes at most one swap per pass after finding the minimum; bubble compares and swaps adjacent pairs, possibly many times per pass.

Worked example

(2 marks) Show the array {6, 3, 8, 2} after the first pass of bubble sort (ascending).

Try it first, then show the solutionSolution

1. (6,3) swap → 3 6 8 2; (6,8) no swap - 1 mark.
2. (8,2) swap → 3 6 2 8 - 1 mark.

Answer: 3 6 2 8

Linear and binary search

Linear search: compare the key with each element from the first to the last until it is found or the array ends. Works on unsorted arrays; slow for large arrays.

int pos = -1;
for (int i = 0; i < n; i++)
if (a[i] == key) { pos = i; break; }

Binary search: works only on a sorted array. Compare the key with the middle element; if equal, found; if the key is smaller, search the left half; if larger, the right half; repeat until found or the half becomes empty.

int low = 0, high = n - 1, pos = -1;
while (low <= high)
{
int mid = (low + high) / 2;
if (a[mid] == key) { pos = mid; break; }
else if (key < a[mid]) high = mid - 1;
else low = mid + 1;
}

Trace in {3, 8, 15, 21, 34, 47, 52, 66} for key 47: low 0, high 7, mid 3 (21 < 47) → low 4; mid 5 (47) → found at index 5.
For key 10: mid 3 (21 > 10) → high 2; mid 1 (8 < 10) → low 2; mid 2 (15 > 10) → high 1; low > high → not found.

For char arrays compare with == and <, as for numbers (not equals()).

Worked example

(2 marks) State two differences between linear and binary search.

Try it first, then show the solutionSolution

1. Linear works on unsorted arrays; binary needs a sorted array - 1 mark.
2. Linear checks elements one by one; binary halves the search range each step, so it is faster for large arrays - 1 mark.

Answer: (1) Linear search works on any array, while binary search works only on a sorted array. (2) Linear search checks elements one by one from the start, while binary search compares with the middle element and halves the range each time, so it is much faster for large arrays.

Double dimensional arrays

A 2D array stores data in rows and columns: int m[][] = new int[3][4]; (3 rows, 4 columns, 12 elements). m[i][j] is the element in row i, column j (both from 0). m.length is the number of rows; m[0].length the number of columns.

Initialisation: int x[][] = {{2, 4}, {6, 8}, {10, 12}}; (3 × 2; x[2][1] = 12).
Input with nested loops:
for (int i = 0; i < 3; i++)
for (int j = 0; j < 3; j++)
m[i][j] = sc.nextInt();

Display in matrix form:
for (int i = 0; i < 3; i++)
{
for (int j = 0; j < 3; j++)
System.out.print(m[i][j] + "\t");
System.out.println();
}

Sums for an n × n matrix:
Row i: add m[i][j] for all j. Column j: add m[i][j] for all i.
Left (main) diagonal: elements where i == j: m[0][0], m[1][1], ...
Right diagonal: elements where i + j == n − 1: m[0][n−1], m[1][n−2], ...

Memory: a char array of 4 rows and 3 columns has 12 elements × 2 bytes = 24 bytes.

Worked example

(2 marks) int x[][] = {{2, 4}, {6, 8}, {10, 12}}; What is x[0][1] + x[2][1]? What is the order of x?

Try it first, then show the solutionSolution

1. x[0][1] = 4; x[2][1] = 12; sum 16 - 1 mark.
2. 3 rows and 2 columns: order 3 × 2 - 1 mark.

Answer: 16; the order is 3 × 2.

Arrays in real life

Each concept, met in an everyday situation. Read the situation, answer the question in your own words, then compare with the model answer. The one-line hook is what to remember.

Topics in this chapter

• Double Dimensional Arrays: Row, Column and Diagonal Sums
• Linear Search and Binary Search Tracing
• Selection Sort and Bubble Sort Pass by Pass

Arrays: syntax and key facts

Index range

0 to a.length − 1; last element a[a.length - 1]

length (array) has no brackets; length() is for String.

Memory

bytes = elements × size of type; 2D elements = rows × columns

int 4, double 8, char 2.

Binary search mid

1. 1.mid = (low + high) / 2; key < a[mid]
2. 2.high = mid − 1; key > a[mid]
3. 3.low = mid + 1

Loop while low <= high.

Diagonals (n × n)

Left: i == j; Right: i + j == n − 1

For 4 × 4: right diagonal m[0][3], m[1][2], m[2][1], m[3][0].

Sort passes

Both sorts	n − 1 passes
Bubble inner loop	j < n − 1 − i
Selection inner loop	j from i + 1

How the board asks Arrays

The question patterns to expect, and the method that works for each.

Tricks to remember Arrays

Short memory aids. Say each one aloud once, then use it on a question the same day.

• Select the min, bubble the max

  Selection sort puts the minimum at the front each pass; bubble sort pushes the maximum to the end.

  Use it when: Telling the sorts apart.

• Binary needs order

  Binary search only on a sorted array; otherwise go linear.

  Use it when: Search questions.

• Right diagonal adds to n − 1

  On the right diagonal, row + column = n − 1.

  Use it when: Diagonal sums.

Arrays: board-level solved questions

Ordered from board level to HOTS. Cover the solution, write your own, then compare step by step.

Case-study practice (Section B style)

Read the situation, then answer every part. In the board paper this is a 10-mark question split into 2 + 3 + 5 marks.

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Where students lose marks in Arrays

The slips examiners see most often, and what to write instead.

• ✗Mistake: Using a different technique from the one asked

  In 2024 some used insertion or bubble sort when selection sort was required; in 2025 some used linear search for binary search.

  ✓Do this: Write exactly the named technique.

  Usually costs: Most of the 15 marks.

• ✗Mistake: Loop bounds 1 to 4 for a 4 × 4 array

  Reported in the 2025 NORM program.

  ✓Do this: Indexes run 0 to 3: for (i = 0; i < 4; i++).

  Usually costs: 2 to 3 marks.

• ✗Mistake: Inverted comparison or incorrect swap

  The 2024 examiners reported both in the selection sort.

  ✓Do this: Swap with a temporary variable: t = a[i]; a[i] = a[min]; a[min] = t;

  Usually costs: 2 to 4 marks.

• ✗Mistake: Confusing length with length()

  The 2025 MCQ on the last index caught many.

  ✓Do this: Array: a.length; String: s.length().

  Usually costs: 1 mark.

• ✗Mistake: Printing both messages in a search

  In 2025 candidates printed 'successful' and 'not successful' together.

  ✓Do this: Use a position/flag variable (or System.exit(0) on success) and print one message after the loop.

  Usually costs: 1 to 2 marks.

• ✗Mistake: Sum not initialised or square root missed

  Reported in the 2025 NORM program.

  ✓Do this: sum = 0 before the loops; Math.sqrt(sum) at the end.

  Usually costs: 2 to 3 marks.

Practice questions on Arrays

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

1. Q1 · 1 mark · MCQBoard level

   int a[] = new int[8]; occupies:
   (a) 8 bytes
   (b) 16 bytes
   (c) 32 bytes
   (d) 64 bytes

   Hint

   8 × 4.

   Answer

   (c) 32 bytes

2. Q2 · 1 mark · MCQBoard level

   String c[] = {"Green", "Red", "Indigo"}; System.out.println(c[2].length()); prints:
   (a) 3
   (b) 6
   (c) 2
   (d) 5

   Hint

   Length of Indigo.

   Answer

   (b) 6

3. Q3 · 2 marks · Very short answerBoard level

   Write a statement to initialise the first six even numbers in a 3 × 2 array named e.

   Hint

   Rows in inner braces.

   Answer

   int e[][] = {{2, 4}, {6, 8}, {10, 12}};

4. Q4 · 2 marks · Very short answerHarder board level

   Give the output:
   int a[] = {5, 2, 8, 1};
   int s = 0;
   for (int i = a.length - 1; i >= 0; i -= 2)
   s = s + a[i];
   System.out.println(s);

   Hint

   Indexes 3 and 1.

   Answer

   3 (a[3] + a[1] = 1 + 2)

5. Q5 · 2 marks · Very short answerHarder board level

   Give the output:
   int m[][] = {{1, 2, 3}, {4, 5, 6}};
   System.out.println(m.length + " " + m[0].length + " " + m[1][2]);

   Hint

   Rows, columns.

   Answer

   2 3 6

6. Q6 · 2 marks · Short answerHOTS / competency

   Why does binary search fail on {9, 3, 7, 1, 5} when searching for 3?

   Hint

   Trace mid.

   Answer

   The array is not sorted. mid = 2 holds 7; 3 < 7 so the search goes left: mid = 0 holds 9; 3 < 9 so high becomes −1 and it stops, reporting not found although 3 is at index 1.

7. Q7 · 15 marks · Long answerHarder board level

   Write a program to accept 10 integers into an array and print the largest and smallest values and their positions (indexes).

   Hint

   Start with a[0].

   Answer

   import java.util.Scanner;
   class MaxMin
   {
   public static void main(String args[])
   {
   Scanner sc = new Scanner(System.in);
   int a[] = new int[10];
   for (int i = 0; i < 10; i++)
   a[i] = sc.nextInt();
   int max = a[0], min = a[0], pmax = 0, pmin = 0;
   for (int i = 1; i < 10; i++)
   {
   if (a[i] > max) { max = a[i]; pmax = i; }
   if (a[i] < min) { min = a[i]; pmin = i; }
   }
   System.out.println("Largest " + max + " at " + pmax);
   System.out.println("Smallest " + min + " at " + pmin);
   }
   }
   Trace with 4 17 9 -3 12 8 17 0 5 6: Largest 17 at 1 (first occurrence, since > not >=); Smallest -3 at 3.

8. Q8 · 15 marks · Long answerHOTS / competency

   Write a program to accept a 4 × 4 integer array and print the square root of the sum of the squares of all its elements.

   Hint

   Initialise sum to 0; Math.sqrt at the end.

   Answer

   import java.util.Scanner;
   class SqrtSum
   {
   public static void main(String args[])
   {
   Scanner sc = new Scanner(System.in);
   int a[][] = new int[4][4];
   long sum = 0;
   for (int i = 0; i < 4; i++)
   for (int j = 0; j < 4; j++)
   {
   a[i][j] = sc.nextInt();
   sum = sum + a[i][j] * a[i][j];
   }
   System.out.println("Result = " + Math.sqrt(sum));
   }
   }
   Trace with all sixteen elements equal to 2: sum = 16 × 4 = 64; Math.sqrt(64) = 8.0 → Result = 8.0.

9. Q9 · 2 marks · Very short answerBoard level

   State one difference between selection sort and bubble sort.

   Hint

   What moves.

   Answer

   Selection sort finds the smallest element of the unsorted part and swaps it into place once per pass, whereas bubble sort compares adjacent elements and swaps them whenever they are out of order, moving the largest to the end each pass.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1. 1.First index of an array?Show answer

   0

2. 2.Which search needs a sorted array?Show answer

   Binary search.

3. 3.Number of elements in int m[3][4]?Show answer

   12

4. 4.Right diagonal condition in n × n?Show answer

   i + j == n − 1

5. 5.Bytes in char c[4][3]?Show answer

   24

Quick revision checklist

The night before the exam: can you do each of these without looking?

• ☐Array: same type, one name, index 0 to length − 1.
• ☐a.length for arrays; s.length() for Strings.
• ☐Selection: select min, swap once per pass; bubble: swap adjacent pairs.
• ☐Linear: any array; binary: sorted array, mid = (low + high) / 2.
• ☐2D: m[i][j]; left diagonal i == j; right i + j == n − 1.
• ☐Sorting and searching only on 1D arrays.

Arrays: frequently asked questions