Tasks Details
hard
Given an array, find all its elements that can become a leader, after increasing by 1 all of the numbers in some segment of a given length.
Task Score
55%
Correctness
100%
Performance
0%
Integers K, M and a non-empty array A consisting of N integers, not bigger than M, are given.
The leader of the array is a value that occurs in more than half of the elements of the array, and the segment of the array is a sequence of consecutive elements of the array.
You can modify A by choosing exactly one segment of length K and increasing by 1 every element within that segment.
The goal is to find all of the numbers that may become a leader after performing exactly one array modification as described above.
Write a function:
fun solution(K: Int, M: Int, A: IntArray): IntArray
that, given integers K and M and an array A consisting of N integers, returns an array of all numbers that can become a leader, after increasing by 1 every element of exactly one segment of A of length K. The returned array should be sorted in ascending order, and if there is no number that can become a leader, you should return an empty array. Moreover, if there are multiple ways of choosing a segment to turn some number into a leader, then this particular number should appear in an output array only once.
For example, given integers K = 3, M = 5 and the following array A:
A[0] = 2 A[1] = 1 A[2] = 3 A[3] = 1 A[4] = 2 A[5] = 2 A[6] = 3the function should return [2, 3]. If we choose segment A[1], A[2], A[3] then we get the following array A:
A[0] = 2 A[1] = 2 A[2] = 4 A[3] = 2 A[4] = 2 A[5] = 2 A[6] = 3and 2 is the leader of this array. If we choose A[3], A[4], A[5] then A will appear as follows:
A[0] = 2 A[1] = 1 A[2] = 3 A[3] = 2 A[4] = 3 A[5] = 3 A[6] = 3and 3 will be the leader.
And, for example, given integers K = 4, M = 2 and the following array:
A[0] = 1 A[1] = 2 A[2] = 2 A[3] = 1 A[4] = 2the function should return [2, 3], because choosing a segment A[0], A[1], A[2], A[3] and A[1], A[2], A[3], A[4] turns 2 and 3 into the leaders, respectively.
Write an efficient algorithm for the following assumptions:
- N and M are integers within the range [1..100,000];
- K is an integer within the range [1..N];
- each element of array A is an integer within the range [1..M].
Copyright 2009–2025 by Codility Limited. All Rights Reserved. Unauthorized copying, publication or disclosure prohibited.
Solution
Programming language used Kotlin
Time spent on task 7 minutes
Notes
not defined yet
Code: 17:56:06 UTC,
java,
autosave
Code: 17:56:11 UTC,
kt,
autosave
Code: 17:58:26 UTC,
kt,
verify,
result: Passed
// you can also use imports, for example:
// import kotlin.math.*
// you can write to stdout for debugging purposes, e.g.
// println("this is a debug message")
fun solution(K: Int, M: Int, A: IntArray): IntArray {
// write your code in Kotlin
var start = 0
val results: ArrayList<Int> = ArrayList()
while (start < A.size - K + 1) {
val newArray = increaseArray(A, start, start + K)
results.add(findOneLeader(M, newArray))
start += 1
}
return removeReduandancy(results.toIntArray())
}
private fun increaseArray(A: IntArray, depart: Int, end: Int): IntArray {
val result = ArrayList<Int>()
A.forEach { a -> result.add(a) }
for (i in depart until end) result[i] = result[i] + 1
return result.toIntArray()
}
private fun findOneLeader(M: Int, A: IntArray): Int {
val nbrs: HashMap<Int, Int?> = HashMap()
for (i in 1 until (M + 2)) nbrs[i] = 0
for (i in A.indices) nbrs[A[i]] = nbrs[A[i]]?.plus(1)
print("\n")
for (i in 1 until (M + 2)) if (nbrs[i]!! > (A.size / 2)) return i
return 0
}
private fun removeReduandancy(A: IntArray): IntArray {
val result = ArrayList<Int>()
A.forEach { a -> if (!result.contains(a) && a > 0) result.add(a) }
result.sort()
return result.toIntArray()
}
Analysis
Code: 18:01:21 UTC,
kt,
verify,
result: Passed
// you can also use imports, for example:
// import kotlin.math.*
// you can write to stdout for debugging purposes, e.g.
// println("this is a debug message")
fun solution(K: Int, M: Int, A: IntArray): IntArray {
// write your code in Kotlin
var start = 0
val results: ArrayList<Int> = ArrayList()
while (start < A.size - K + 1) {
val newArray = increaseArray(A, start, start + K)
results.add(findOneLeader(M, newArray))
start += 1
}
return removeReduandancy(results.toIntArray())
}
private fun increaseArray(A: IntArray, depart: Int, end: Int): IntArray {
val result = ArrayList<Int>()
A.forEach { a -> result.add(a) }
for (i in depart until end) result[i] = result[i] + 1
return result.toIntArray()
}
private fun findOneLeader(M: Int, A: IntArray): Int {
val nbrs: HashMap<Int, Int?> = HashMap()
for (i in 1 until (M + 2)) nbrs[i] = 0
for (i in A.indices) nbrs[A[i]] = nbrs[A[i]]?.plus(1)
print("\n")
for (i in 1 until (M + 2)) if (nbrs[i]!! > (A.size / 2)) return i
return 0
}
private fun removeReduandancy(A: IntArray): IntArray {
val result = ArrayList<Int>()
A.forEach { a -> if (!result.contains(a) && a > 0) result.add(a) }
result.sort()
return result.toIntArray()
}
Analysis
Code: 18:01:48 UTC,
kt,
autosave
// you can also use imports, for example:
// import kotlin.math.*
// you can write to stdout for debugging purposes, e.g.
// println("this is a debug message")
fun solution(K: Int, M: Int, A: IntArray): IntArray {
// write your code in Kotlin
var start = 0
val results: ArrayList<Int> = ArrayList()
while (start < A.size - K + 1) {
val newArray = increaseArray(A, start, start + K)
results.add(findOneLeader(M, newArray))
start += 1
}
return removeReduandancy(results.toIntArray())
}
private fun increaseArray(A: IntArray, depart: Int, end: Int): IntArray {
val result = ArrayList<Int>()
A.forEach { a -> result.add(a) }
for (i in depart until end) result[i] = result[i] + 1
return result.toIntArray()
}
private fun findOneLeader(M: Int, A: IntArray): Int {
val nbrs: HashMap<Int, Int?> = HashMap()
for (i in 1 until (M + 2)) nbrs[i] = 0
for (i in A.indices) nbrs[A[i]] = nbrs[A[i]]?.plus(1)
for (i in 1 until (M + 2)) if (nbrs[i]!! > (A.size / 2)) return i
return 0
}
private fun removeReduandancy(A: IntArray): IntArray {
val result = ArrayList<Int>()
A.forEach { a -> if (!result.contains(a) && a > 0) result.add(a) }
result.sort()
return result.toIntArray()
}
Code: 18:01:50 UTC,
kt,
verify,
result: Passed
// you can also use imports, for example:
// import kotlin.math.*
// you can write to stdout for debugging purposes, e.g.
// println("this is a debug message")
fun solution(K: Int, M: Int, A: IntArray): IntArray {
// write your code in Kotlin
var start = 0
val results: ArrayList<Int> = ArrayList()
while (start < A.size - K + 1) {
val newArray = increaseArray(A, start, start + K)
results.add(findOneLeader(M, newArray))
start += 1
}
return removeReduandancy(results.toIntArray())
}
private fun increaseArray(A: IntArray, depart: Int, end: Int): IntArray {
val result = ArrayList<Int>()
A.forEach { a -> result.add(a) }
for (i in depart until end) result[i] = result[i] + 1
return result.toIntArray()
}
private fun findOneLeader(M: Int, A: IntArray): Int {
val nbrs: HashMap<Int, Int?> = HashMap()
for (i in 1 until (M + 2)) nbrs[i] = 0
for (i in A.indices) nbrs[A[i]] = nbrs[A[i]]?.plus(1)
for (i in 1 until (M + 2)) if (nbrs[i]!! > (A.size / 2)) return i
return 0
}
private fun removeReduandancy(A: IntArray): IntArray {
val result = ArrayList<Int>()
A.forEach { a -> if (!result.contains(a) && a > 0) result.add(a) }
result.sort()
return result.toIntArray()
}
Analysis
Code: 18:02:03 UTC,
kt,
verify,
result: Passed
// you can also use imports, for example:
// import kotlin.math.*
// you can write to stdout for debugging purposes, e.g.
// println("this is a debug message")
fun solution(K: Int, M: Int, A: IntArray): IntArray {
// write your code in Kotlin
var start = 0
val results: ArrayList<Int> = ArrayList()
while (start < A.size - K + 1) {
val newArray = increaseArray(A, start, start + K)
results.add(findOneLeader(M, newArray))
start += 1
}
return removeReduandancy(results.toIntArray())
}
private fun increaseArray(A: IntArray, depart: Int, end: Int): IntArray {
val result = ArrayList<Int>()
A.forEach { a -> result.add(a) }
for (i in depart until end) result[i] = result[i] + 1
return result.toIntArray()
}
private fun findOneLeader(M: Int, A: IntArray): Int {
val nbrs: HashMap<Int, Int?> = HashMap()
for (i in 1 until (M + 2)) nbrs[i] = 0
for (i in A.indices) nbrs[A[i]] = nbrs[A[i]]?.plus(1)
for (i in 1 until (M + 2)) if (nbrs[i]!! > (A.size / 2)) return i
return 0
}
private fun removeReduandancy(A: IntArray): IntArray {
val result = ArrayList<Int>()
A.forEach { a -> if (!result.contains(a) && a > 0) result.add(a) }
result.sort()
return result.toIntArray()
}
Analysis
Code: 18:02:15 UTC,
kt,
final,
score:
55
// you can also use imports, for example:
// import kotlin.math.*
// you can write to stdout for debugging purposes, e.g.
// println("this is a debug message")
fun solution(K: Int, M: Int, A: IntArray): IntArray {
// write your code in Kotlin
var start = 0
val results: ArrayList<Int> = ArrayList()
while (start < A.size - K + 1) {
val newArray = increaseArray(A, start, start + K)
results.add(findOneLeader(M, newArray))
start += 1
}
return removeReduandancy(results.toIntArray())
}
private fun increaseArray(A: IntArray, depart: Int, end: Int): IntArray {
val result = ArrayList<Int>()
A.forEach { a -> result.add(a) }
for (i in depart until end) result[i] = result[i] + 1
return result.toIntArray()
}
private fun findOneLeader(M: Int, A: IntArray): Int {
val nbrs: HashMap<Int, Int?> = HashMap()
for (i in 1 until (M + 2)) nbrs[i] = 0
for (i in A.indices) nbrs[A[i]] = nbrs[A[i]]?.plus(1)
for (i in 1 until (M + 2)) if (nbrs[i]!! > (A.size / 2)) return i
return 0
}
private fun removeReduandancy(A: IntArray): IntArray {
val result = ArrayList<Int>()
A.forEach { a -> if (!result.contains(a) && a > 0) result.add(a) }
result.sort()
return result.toIntArray()
}Analysis summary
The following issues have been detected: timeout errors.
Analysis
Detected time complexity:
O(N*(N+M)) or O(N**2)
expand all
Correctness tests
1.
0.020 s
OK
2.
0.024 s
OK
3.
0.020 s
OK
4.
0.024 s
OK
5.
0.324 s
OK
6.
0.288 s
OK
1.
0.020 s
OK
2.
0.024 s
OK
3.
0.024 s
OK
4.
0.020 s
OK
1.
0.024 s
OK
2.
0.020 s
OK
1.
0.024 s
OK
2.
0.024 s
OK
1.
0.020 s
OK
2.
0.024 s
OK
3.
0.024 s
OK
expand all
Performance tests
medium_tests1
medium tests (N = 10000, M = 100)
medium tests (N = 10000, M = 100)
✘
TIMEOUT ERROR
running time: 2.020 sec., time limit: 0.128 sec.
running time: 2.020 sec., time limit: 0.128 sec.
1.
2.020 s
TIMEOUT ERROR,
running time: 2.020 sec., time limit: 0.128 sec.
2.
2.548 s
TIMEOUT ERROR,
running time: 2.548 sec., time limit: 0.144 sec.
3.
2.416 s
TIMEOUT ERROR,
running time: 2.416 sec., time limit: 0.128 sec.
medium_tests2
medium tests(N >= 20000, M=30000)
medium tests(N >= 20000, M=30000)
✘
TIMEOUT ERROR
Killed. Hard limit reached: 6.000 sec.
Killed. Hard limit reached: 6.000 sec.
1.
6.000 s
TIMEOUT ERROR,
Killed. Hard limit reached: 6.000 sec.
2.
6.000 s
TIMEOUT ERROR,
Killed. Hard limit reached: 6.000 sec.
3.
6.000 s
TIMEOUT ERROR,
Killed. Hard limit reached: 6.000 sec.
1.
6.000 s
TIMEOUT ERROR,
Killed. Hard limit reached: 6.000 sec.
2.
6.000 s
TIMEOUT ERROR,
Killed. Hard limit reached: 6.000 sec.
3.
6.000 s
TIMEOUT ERROR,
Killed. Hard limit reached: 6.000 sec.
1.
0.584 s
TIMEOUT ERROR,
running time: 0.584 sec., time limit: 0.256 sec.
2.
6.000 s
TIMEOUT ERROR,
Killed. Hard limit reached: 6.000 sec.