Tasks Details
medium
Divide array A into K blocks and minimize the largest sum of any block.
Task Score
100%
Correctness
100%
Performance
100%
You are given integers K, M and a non-empty array A consisting of N integers. Every element of the array is not greater than M.
You should divide this array into K blocks of consecutive elements. The size of the block is any integer between 0 and N. Every element of the array should belong to some block.
The sum of the block from X to Y equals A[X] + A[X + 1] + ... + A[Y]. The sum of empty block equals 0.
The large sum is the maximal sum of any block.
For example, you are given integers K = 3, M = 5 and array A such that:
A[0] = 2 A[1] = 1 A[2] = 5 A[3] = 1 A[4] = 2 A[5] = 2 A[6] = 2The array can be divided, for example, into the following blocks:
- [2, 1, 5, 1, 2, 2, 2], [], [] with a large sum of 15;
- [2], [1, 5, 1, 2], [2, 2] with a large sum of 9;
- [2, 1, 5], [], [1, 2, 2, 2] with a large sum of 8;
- [2, 1], [5, 1], [2, 2, 2] with a large sum of 6.
The goal is to minimize the large sum. In the above example, 6 is the minimal large sum.
Write a function:
class Solution { public int solution(int K, int M, int[] A); }
that, given integers K, M and a non-empty array A consisting of N integers, returns the minimal large sum.
For example, given K = 3, M = 5 and array A such that:
A[0] = 2 A[1] = 1 A[2] = 5 A[3] = 1 A[4] = 2 A[5] = 2 A[6] = 2the function should return 6, as explained above.
Write an efficient algorithm for the following assumptions:
- N and K are integers within the range [1..100,000];
- M is an integer within the range [0..10,000];
- each element of array A is an integer within the range [0..M].
Copyright 2009–2025 by Codility Limited. All Rights Reserved. Unauthorized copying, publication or disclosure prohibited.
Solution
Programming language used Java 21
Time spent on task 1 minutes
Notes
not defined yet
Code: 17:13:18 UTC,
java,
verify,
result: Passed
12345678910111213141516171819202122232425262728293031323334353637383940414243444546474849505152535455565758596061626364656667
// 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 K, int M, int[] A) {
int min = 0;
int max = 0;
for (int i = 0; i < A.length; i++) {
max += A[i];
min = Math.max(min, A[i]);
}
if (K == 1)
return max;
if (K >= A.length)
return min;
int result = min;
while (min <= max) {
int mid = (min + max) / 2;
if (check(mid, K, A)) {
max = mid - 1;
result = mid;
} else {
min = mid + 1;
}
}
return result;
}
private boolean check(int mid, int k, int[] a) {
int sum = 0;
for (int i = 0; i < a.length; i++) {
sum += a[i];
if (sum > mid) {
sum = a[i];
k--;
}
if (k == 0)
return false;
}
return true;
}
}
Analysis
Code: 17:13:23 UTC,
java,
verify,
result: Passed
12345678910111213141516171819202122232425262728293031323334353637383940414243444546474849505152535455565758596061626364656667
// 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 K, int M, int[] A) {
int min = 0;
int max = 0;
for (int i = 0; i < A.length; i++) {
max += A[i];
min = Math.max(min, A[i]);
}
if (K == 1)
return max;
if (K >= A.length)
return min;
int result = min;
while (min <= max) {
int mid = (min + max) / 2;
if (check(mid, K, A)) {
max = mid - 1;
result = mid;
} else {
min = mid + 1;
}
}
return result;
}
private boolean check(int mid, int k, int[] a) {
int sum = 0;
for (int i = 0; i < a.length; i++) {
sum += a[i];
if (sum > mid) {
sum = a[i];
k--;
}
if (k == 0)
return false;
}
return true;
}
}
Analysis
Code: 17:13:27 UTC,
java,
final,
score:
100
12345678910111213141516171819202122232425262728293031323334353637383940414243444546474849505152535455565758596061626364656667
// 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 K, int M, int[] A) {
int min = 0;
int max = 0;
for (int i = 0; i < A.length; i++) {
max += A[i];
min = Math.max(min, A[i]);
}
if (K == 1)
return max;
if (K >= A.length)
return min;
int result = min;
while (min <= max) {
int mid = (min + max) / 2;
if (check(mid, K, A)) {
max = mid - 1;
result = mid;
} else {
min = mid + 1;
}
}
return result;
}
private boolean check(int mid, int k, int[] a) {
int sum = 0;
for (int i = 0; i < a.length; i++) {
sum += a[i];
if (sum > mid) {
sum = a[i];
k--;
}
if (k == 0)
return false;
}
return true;
}
}
Analysis summary
The solution obtained perfect score.
Analysis
Detected time complexity:
O(N*log(N+M))
expand all
Correctness tests
1.
0.008 s
OK
2.
0.008 s
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3.
0.008 s
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4.
0.008 s
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1.
0.008 s
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2.
0.008 s
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3.
0.008 s
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4.
0.008 s
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5.
0.008 s
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6.
0.008 s
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1.
0.008 s
OK
2.
0.008 s
OK
3.
0.008 s
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4.
0.004 s
OK
1.
0.008 s
OK
2.
0.008 s
OK
3.
0.008 s
OK
1.
0.008 s
OK
2.
0.008 s
OK
3.
0.008 s
OK
1.
0.008 s
OK
2.
0.004 s
OK
3.
0.008 s
OK
4.
0.008 s
OK
expand all
Performance tests
1.
0.008 s
OK
2.
0.008 s
OK
3.
0.008 s
OK
4.
0.008 s
OK
1.
0.032 s
OK
2.
0.036 s
OK
3.
0.036 s
OK
4.
0.040 s
OK
1.
0.048 s
OK
2.
0.052 s
OK
3.
0.052 s
OK
4.
0.052 s
OK
1.
0.288 s
OK
2.
0.292 s
OK
3.
0.292 s
OK
4.
0.292 s
OK
1.
0.160 s
OK
2.
0.188 s
OK
3.
0.196 s
OK
4.
0.196 s
OK
1.
0.256 s
OK
2.
0.312 s
OK
3.
0.316 s
OK
4.
0.256 s
OK