Tasks Details
easy
1.
Dominator
Find an index of an array such that its value occurs at more than half of indices in the array.
Task Score
100%
Correctness
100%
Performance
100%
An array A consisting of N integers is given. The dominator of array A is the value that occurs in more than half of the elements of A.
For example, consider array A such that
A[0] = 3 A[1] = 4 A[2] = 3 A[3] = 2 A[4] = 3 A[5] = -1 A[6] = 3 A[7] = 3The dominator of A is 3 because it occurs in 5 out of 8 elements of A (namely in those with indices 0, 2, 4, 6 and 7) and 5 is more than a half of 8.
Write a function
class Solution { public int solution(int[] A); }
that, given an array A consisting of N integers, returns index of any element of array A in which the dominator of A occurs. The function should return −1 if array A does not have a dominator.
For example, given array A such that
A[0] = 3 A[1] = 4 A[2] = 3 A[3] = 2 A[4] = 3 A[5] = -1 A[6] = 3 A[7] = 3the function may return 0, 2, 4, 6 or 7, as explained above.
Write an efficient algorithm for the following assumptions:
- N is an integer within the range [0..100,000];
- each element of array A is an integer within the range [−2,147,483,648..2,147,483,647].
Copyright 2009–2024 by Codility Limited. All Rights Reserved. Unauthorized copying, publication or disclosure prohibited.
Solution
Programming language used Java 8
Time spent on task 58 minutes
Notes
not defined yet
Task timeline
Code: 11:30:26 UTC,
java,
autosave
Code: 11:56:44 UTC,
cpp,
autosave
Code: 11:56:48 UTC,
java,
autosave
Code: 12:27:05 UTC,
java,
autosave
// you can also use imports, for example:
// import java.util.*;
// you can write to stdout for debugging purposes, e.g.
// System.out.println("this is a debug message");
class Solution {
public int solution(int[] A) {
int N = A.length;
if (N == 0) return -1;
Stack<Integer> stack = new Stack<>();
for (int num : A) {
if (stack.isEmpty()) {
stack.push(num);
continue;
}
if (stack.peek() == num) stack.push(num);
else stack.pop();
}
int candidate = -1;
int leaderIndex = -1;
int count = 0;
if (stack.size() > 0) candidate = stack.peek();
for (int i = 0; i < N; i++) {
if (candidate == A[i]) count++;
if (count > (N / 2)) {
leaderIndex = i;
break;
}
}
return leaderIndex;
}
}
Code: 12:27:13 UTC,
java,
verify,
result: Passed
// you can also use imports, for example:
// import java.util.*;
// you can write to stdout for debugging purposes, e.g.
// System.out.println("this is a debug message");
import java.util.Stack;
class Solution {
public int solution(int[] A) {
int N = A.length;
if (N == 0) return -1;
Stack<Integer> stack = new Stack<>();
for (int num : A) {
if (stack.isEmpty()) {
stack.push(num);
continue;
}
if (stack.peek() == num) stack.push(num);
else stack.pop();
}
int candidate = -1;
int leaderIndex = -1;
int count = 0;
if (stack.size() > 0) candidate = stack.peek();
for (int i = 0; i < N; i++) {
if (candidate == A[i]) count++;
if (count > (N / 2)) {
leaderIndex = i;
break;
}
}
return leaderIndex;
}
}
Analysis
Code: 12:27:22 UTC,
java,
autosave
// you can also use imports, for example:
// import java.util.*;
// you can write to stdout for debugging purposes, e.g.
// System.out.println("this is a debug message");
import java.util.Stack;
class Solution {
public int solution(int[] A) {
int N = A.length;
if (N == 0) return -1;
Stack<Integer> stack = new Stack<>();
for (int num : A) {
if (stack.isEmpty()) {
stack.push(num);
continue;
}
if (stack.peek() == num) stack.push(num);
else stack.pop();
}
int candidate = -1;
int leaderIndex = -1;
int count = 0;
if (stack.size() > 0) candidate = stack.peek();
for (int i = 0; i < N; i++) {
if (candidate == A[i]) count++;
if (count > (N / 2)) {
leaderIndex = i;
break;
}
}
return leaderIndex;
}
}
Code: 12:27:25 UTC,
java,
verify,
result: Passed
// you can also use imports, for example:
// import java.util.*;
// you can write to stdout for debugging purposes, e.g.
// System.out.println("this is a debug message");
import java.util.Stack;
class Solution {
public int solution(int[] A) {
int N = A.length;
if (N == 0) return -1;
Stack<Integer> stack = new Stack<>();
for (int num : A) {
if (stack.isEmpty()) {
stack.push(num);
continue;
}
if (stack.peek() == num) stack.push(num);
else stack.pop();
}
int candidate = -1;
int leaderIndex = -1;
int count = 0;
if (stack.size() > 0) candidate = stack.peek();
for (int i = 0; i < N; i++) {
if (candidate == A[i]) count++;
if (count > (N / 2)) {
leaderIndex = i;
break;
}
}
return leaderIndex;
}
}
Analysis
Code: 12:27:29 UTC,
java,
final,
score:
100
// you can also use imports, for example:
// import java.util.*;
// you can write to stdout for debugging purposes, e.g.
// System.out.println("this is a debug message");
import java.util.Stack;
class Solution {
public int solution(int[] A) {
int N = A.length;
if (N == 0) return -1;
Stack<Integer> stack = new Stack<>();
for (int num : A) {
if (stack.isEmpty()) {
stack.push(num);
continue;
}
if (stack.peek() == num) stack.push(num);
else stack.pop();
}
int candidate = -1;
int leaderIndex = -1;
int count = 0;
if (stack.size() > 0) candidate = stack.peek();
for (int i = 0; i < N; i++) {
if (candidate == A[i]) count++;
if (count > (N / 2)) {
leaderIndex = i;
break;
}
}
return leaderIndex;
}
}Analysis summary
The solution obtained perfect score.
Analysis
Detected time complexity:
O(N*log(N)) or O(N)
expand all
Correctness tests
1.
0.004 s
OK
2.
0.004 s
OK
1.
0.004 s
OK
2.
0.008 s
OK
1.
0.008 s
OK
2.
0.004 s
OK
1.
0.004 s
OK
1.
0.004 s
OK
2.
0.008 s
OK
1.
0.004 s
OK
2.
0.008 s
OK
1.
0.004 s
OK
2.
0.004 s
OK
1.
0.004 s
OK
2.
0.004 s
OK