Test results - Codility

Tasks Details

hard

1. ABSwaps

Calculate the minimum required number of swaps of neighbouring letters in a string built from the letters 'a' and 'b', after which the string would contain both "aaa" and "bbb" as substrings.

75%

100%

60%

You are given a string S consisting of letters 'a' and 'b'. The task is to rearrange letters in S so that it contains three consecutive letters 'a' and three consecutive letters 'b'. What is the minimum necessary number of swaps of neighbouring letters to achieve this?

Write a function:

class Solution { public int solution(String S); }

that, given a string S of length N, returns the number of swaps after which S would contain "aaa" and "bbb" as substrings. If it's not possible to rearrange the letters in such a way, the function should return −1.

Examples:

1. Given S = "ababab", the function should return 3. The sequence of swaps could be as follows: ababab → abaabb → aababb → aaabbb.

2. Given S = "abbbbaa", the function should return 4. The sequence of four swaps is: abbbbaa → babbbaa → bbabbaa → bbbabaa → bbbbaaa.

3. Given S = "bbbababaaab", the function should return 0. S already contains both "aaa" and "bbb" as substrings.

4. Given S = "abbabb", the function should return −1. It is not possible to obtain the required result from S as there are only two letters 'a'.

Write an efficient algorithm for the following assumptions:

N is an integer within the range [1..100,000];

string S is made only of the characters 'a' and/or 'b'.

Solution

Programming language used Java 21

Time spent on task 1 minutes

Notes

not defined yet

Code: 05:58:58 UTC, java, verify, result: Passed

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class Solution {
	// B2
	final static int BIG_NUMBER = 1000001;

	class MinNode {
		int count = 0;
		int type;
		int minTo3[] = new int[] { BIG_NUMBER, BIG_NUMBER };

		int max[] = new int[] { 0, 0 };
		int min[] = new int[] { BIG_NUMBER, BIG_NUMBER };
		int totalCount[] = new int[] { 0, 0 };
		MinNode left, right;

		public void initTree(int from, int to) {
			int m = (from + to) / 2;
			if (from != m) {
				left = new MinNode();
				left.initTree(from, m);
			}
			if (m + 1 != to) {
				right = new MinNode();
				right.initTree(m + 1, to);
			}
		}

		public void initNode(MinNode mn, int i, int from, int to) {
			int m = (from + to) / 2;
			if (from == m && from == i)
				left = mn;
			else if (m + 1 == to && to == i)
				right = mn;
			else if (from != m && i >= from && i <= m)
				left.initNode(mn, i, from, m);
			else if (to != m + 1 && i >= m + 1 && i <= to)
				right.initNode(mn, i, m + 1, to);
		}

		public void update(int updateFrom, int updateTo, int from, int to) {
			if (from == to) {
				totalCount[type] = count;
				max[type] = count;
				min[type] = count;
				return;
			}
			int m = (from + to) / 2;
			if (updateFrom <= m && updateTo >= from)
				left.update(updateFrom, updateTo, from, m);
			if (updateFrom <= to && updateTo >= m + 1)
				right.update(updateFrom, updateTo, m + 1, to);

			totalCount[0] = left.totalCount[0] + right.totalCount[0];
			totalCount[1] += left.totalCount[1] + right.totalCount[1];
			max[0] = Math.max(right.max[0], left.max[0]);
			max[1] = Math.max(right.max[1], left.max[1]);
			min[0] = Math.min(right.min[0], left.min[0]);
			min[1] = Math.min(right.min[1], left.min[1]);
			minTo3[0] = Math.min(right.minTo3[0], left.minTo3[0]);
			minTo3[1] = Math.min(right.minTo3[1], left.minTo3[1]);
		}
	}

	MinNode[] N;
	int nodeCount = 0;
	MinNode Root = new MinNode();

	public int solution(String S) {
		char[] s = S.toCharArray();
		N = new MinNode[s.length];

		MinNode currentNode = new MinNode();
		N[nodeCount] = currentNode;
		currentNode.count = 1;
		currentNode.type = s[0] == 'a' ? 0 : 1;

		for (int i = 1; i < s.length; i++) {
			int ct = s[i] == 'a' ? 0 : 1;
			if (ct == currentNode.type)
				currentNode.count++;
			else {
				currentNode = new MinNode();
				currentNode.count = 1;
				currentNode.type = ct;
				N[++nodeCount] = currentNode;
			}
		}
		nodeCount++;
		if (nodeCount < 2)
			return -1;

		Root.initTree(0, nodeCount - 1);
		boolean has2112 = false;

		for (int i = 0; i < nodeCount; i++) {
			if (!has2112 && i < nodeCount - 3) {
				has2112 = ((N[i].count > 1) && (N[i + 1].count == 1) && (N[i + 2].count == 1) && (N[i + 3].count > 1));
			}
			updateMinTo3(i);
			Root.initNode(N[i], i, 0, nodeCount - 1);
		}
		Root.update(0, nodeCount - 1, 0, nodeCount - 1);

		if (Root.totalCount[0] < 3 || Root.totalCount[1] < 3)
			return -1;
		if (Root.max[0] >= 3 && Root.max[1] >= 3)
			return 0;
		if (has2112)
			return 1;
		if (Root.max[0] == 1 && Root.max[1] == 1)
			return 3;
		if (Root.max[0] == 2 && Root.max[1] == 2 && Root.min[0] == 2 && Root.min[1] == 2)
			return 3;
		if (Root.max[0] >= 3)
			return Root.minTo3[1];
		if (Root.max[1] >= 3)
			return Root.minTo3[0];

		int m = BIG_NUMBER;

		for (int i = 0; i < nodeCount && m > 2; i++) {
			MinNode mn = N[i];
			if (mn.count == 2) {
				if (i > 1) {
					m = testSwapNode(i - 2, i, -1, N[i - 1], m);
					m = testSwapNode(i - 2, i, 2, N[i - 1], m);
				}
				if (i < nodeCount - 2) {
					m = testSwapNode(i, i + 2, 1, N[i + 1], m);
					m = testSwapNode(i, i + 2, -2, N[i + 1], m);
				}
			}
		}

		return m;
	}

	int testSwapNode(int base, int target, int testValue, MinNode using, int m) {
		int from = (base == 0 ? 0 : base - 1);
		int to = (target == nodeCount - 1 ? nodeCount - 1 : target + 1);

		N[base].count += testValue;
		N[target].count -= testValue;

		for (int i = from; i <= to; i++)
			updateMinTo3(i);

		Root.update(from, to, 0, nodeCount - 1);

		m = Math.min(m, using.count * Math.abs(testValue) + Root.minTo3[using.type]);

		N[base].count -= testValue;
		N[target].count += testValue;
		for (int i = from; i <= to; i++)
			updateMinTo3(i);

		Root.update(from, to, 0, nodeCount - 1);
		return m;
	}

	private void updateMinTo3(int i) {
		MinNode n = N[i];

		int l = (i > 1) ? N[i - 1].count : BIG_NUMBER;
		int r = (i < nodeCount - 2) ? N[i + 1].count : BIG_NUMBER;

		n.minTo3[n.type] = BIG_NUMBER;
		if (n.count == 1 && i > 1 && i < nodeCount - 2)
			n.minTo3[n.type] = l + r;
		else if (n.count == 2)
			n.minTo3[n.type] = Math.min(l, r);
		else if (n.count > 2)
			n.minTo3[n.type] = 0;
	}
}

Analysis

▶

example1
First example test.

✔

▶

example2
Second example test.

✔

▶

example3
Third example test.

✔

▶

example4
Fourth example test.

✔

Code: 05:59:05 UTC, java, verify, result: Passed

class Solution {
	// B2
	final static int BIG_NUMBER = 1000001;

	class MinNode {
		int count = 0;
		int type;
		int minTo3[] = new int[] { BIG_NUMBER, BIG_NUMBER };

		int max[] = new int[] { 0, 0 };
		int min[] = new int[] { BIG_NUMBER, BIG_NUMBER };
		int totalCount[] = new int[] { 0, 0 };
		MinNode left, right;

		public void initTree(int from, int to) {
			int m = (from + to) / 2;
			if (from != m) {
				left = new MinNode();
				left.initTree(from, m);
			}
			if (m + 1 != to) {
				right = new MinNode();
				right.initTree(m + 1, to);
			}
		}

		public void initNode(MinNode mn, int i, int from, int to) {
			int m = (from + to) / 2;
			if (from == m && from == i)
				left = mn;
			else if (m + 1 == to && to == i)
				right = mn;
			else if (from != m && i >= from && i <= m)
				left.initNode(mn, i, from, m);
			else if (to != m + 1 && i >= m + 1 && i <= to)
				right.initNode(mn, i, m + 1, to);
		}

		public void update(int updateFrom, int updateTo, int from, int to) {
			if (from == to) {
				totalCount[type] = count;
				max[type] = count;
				min[type] = count;
				return;
			}
			int m = (from + to) / 2;
			if (updateFrom <= m && updateTo >= from)
				left.update(updateFrom, updateTo, from, m);
			if (updateFrom <= to && updateTo >= m + 1)
				right.update(updateFrom, updateTo, m + 1, to);

			totalCount[0] = left.totalCount[0] + right.totalCount[0];
			totalCount[1] += left.totalCount[1] + right.totalCount[1];
			max[0] = Math.max(right.max[0], left.max[0]);
			max[1] = Math.max(right.max[1], left.max[1]);
			min[0] = Math.min(right.min[0], left.min[0]);
			min[1] = Math.min(right.min[1], left.min[1]);
			minTo3[0] = Math.min(right.minTo3[0], left.minTo3[0]);
			minTo3[1] = Math.min(right.minTo3[1], left.minTo3[1]);
		}
	}

	MinNode[] N;
	int nodeCount = 0;
	MinNode Root = new MinNode();

	public int solution(String S) {
		char[] s = S.toCharArray();
		N = new MinNode[s.length];

		MinNode currentNode = new MinNode();
		N[nodeCount] = currentNode;
		currentNode.count = 1;
		currentNode.type = s[0] == 'a' ? 0 : 1;

		for (int i = 1; i < s.length; i++) {
			int ct = s[i] == 'a' ? 0 : 1;
			if (ct == currentNode.type)
				currentNode.count++;
			else {
				currentNode = new MinNode();
				currentNode.count = 1;
				currentNode.type = ct;
				N[++nodeCount] = currentNode;
			}
		}
		nodeCount++;
		if (nodeCount < 2)
			return -1;

		Root.initTree(0, nodeCount - 1);
		boolean has2112 = false;

		for (int i = 0; i < nodeCount; i++) {
			if (!has2112 && i < nodeCount - 3) {
				has2112 = ((N[i].count > 1) && (N[i + 1].count == 1) && (N[i + 2].count == 1) && (N[i + 3].count > 1));
			}
			updateMinTo3(i);
			Root.initNode(N[i], i, 0, nodeCount - 1);
		}
		Root.update(0, nodeCount - 1, 0, nodeCount - 1);

		if (Root.totalCount[0] < 3 || Root.totalCount[1] < 3)
			return -1;
		if (Root.max[0] >= 3 && Root.max[1] >= 3)
			return 0;
		if (has2112)
			return 1;
		if (Root.max[0] == 1 && Root.max[1] == 1)
			return 3;
		if (Root.max[0] == 2 && Root.max[1] == 2 && Root.min[0] == 2 && Root.min[1] == 2)
			return 3;
		if (Root.max[0] >= 3)
			return Root.minTo3[1];
		if (Root.max[1] >= 3)
			return Root.minTo3[0];

		int m = BIG_NUMBER;

		for (int i = 0; i < nodeCount && m > 2; i++) {
			MinNode mn = N[i];
			if (mn.count == 2) {
				if (i > 1) {
					m = testSwapNode(i - 2, i, -1, N[i - 1], m);
					m = testSwapNode(i - 2, i, 2, N[i - 1], m);
				}
				if (i < nodeCount - 2) {
					m = testSwapNode(i, i + 2, 1, N[i + 1], m);
					m = testSwapNode(i, i + 2, -2, N[i + 1], m);
				}
			}
		}

		return m;
	}

	int testSwapNode(int base, int target, int testValue, MinNode using, int m) {
		int from = (base == 0 ? 0 : base - 1);
		int to = (target == nodeCount - 1 ? nodeCount - 1 : target + 1);

		N[base].count += testValue;
		N[target].count -= testValue;

		for (int i = from; i <= to; i++)
			updateMinTo3(i);

		Root.update(from, to, 0, nodeCount - 1);

		m = Math.min(m, using.count * Math.abs(testValue) + Root.minTo3[using.type]);

		N[base].count -= testValue;
		N[target].count += testValue;
		for (int i = from; i <= to; i++)
			updateMinTo3(i);

		Root.update(from, to, 0, nodeCount - 1);
		return m;
	}

	private void updateMinTo3(int i) {
		MinNode n = N[i];

		int l = (i > 1) ? N[i - 1].count : BIG_NUMBER;
		int r = (i < nodeCount - 2) ? N[i + 1].count : BIG_NUMBER;

		n.minTo3[n.type] = BIG_NUMBER;
		if (n.count == 1 && i > 1 && i < nodeCount - 2)
			n.minTo3[n.type] = l + r;
		else if (n.count == 2)
			n.minTo3[n.type] = Math.min(l, r);
		else if (n.count > 2)
			n.minTo3[n.type] = 0;
	}
}

Analysis

▶

example1
First example test.

✔

▶

example2
Second example test.

✔

▶

example3
Third example test.

✔

▶

example4
Fourth example test.

✔

Code: 05:59:11 UTC, java, final, score: 75

class Solution {
	// B2
	final static int BIG_NUMBER = 1000001;

	class MinNode {
		int count = 0;
		int type;
		int minTo3[] = new int[] { BIG_NUMBER, BIG_NUMBER };

		int max[] = new int[] { 0, 0 };
		int min[] = new int[] { BIG_NUMBER, BIG_NUMBER };
		int totalCount[] = new int[] { 0, 0 };
		MinNode left, right;

		public void initTree(int from, int to) {
			int m = (from + to) / 2;
			if (from != m) {
				left = new MinNode();
				left.initTree(from, m);
			}
			if (m + 1 != to) {
				right = new MinNode();
				right.initTree(m + 1, to);
			}
		}

		public void initNode(MinNode mn, int i, int from, int to) {
			int m = (from + to) / 2;
			if (from == m && from == i)
				left = mn;
			else if (m + 1 == to && to == i)
				right = mn;
			else if (from != m && i >= from && i <= m)
				left.initNode(mn, i, from, m);
			else if (to != m + 1 && i >= m + 1 && i <= to)
				right.initNode(mn, i, m + 1, to);
		}

		public void update(int updateFrom, int updateTo, int from, int to) {
			if (from == to) {
				totalCount[type] = count;
				max[type] = count;
				min[type] = count;
				return;
			}
			int m = (from + to) / 2;
			if (updateFrom <= m && updateTo >= from)
				left.update(updateFrom, updateTo, from, m);
			if (updateFrom <= to && updateTo >= m + 1)
				right.update(updateFrom, updateTo, m + 1, to);

			totalCount[0] = left.totalCount[0] + right.totalCount[0];
			totalCount[1] += left.totalCount[1] + right.totalCount[1];
			max[0] = Math.max(right.max[0], left.max[0]);
			max[1] = Math.max(right.max[1], left.max[1]);
			min[0] = Math.min(right.min[0], left.min[0]);
			min[1] = Math.min(right.min[1], left.min[1]);
			minTo3[0] = Math.min(right.minTo3[0], left.minTo3[0]);
			minTo3[1] = Math.min(right.minTo3[1], left.minTo3[1]);
		}
	}

	MinNode[] N;
	int nodeCount = 0;
	MinNode Root = new MinNode();

	public int solution(String S) {
		char[] s = S.toCharArray();
		N = new MinNode[s.length];

		MinNode currentNode = new MinNode();
		N[nodeCount] = currentNode;
		currentNode.count = 1;
		currentNode.type = s[0] == 'a' ? 0 : 1;

		for (int i = 1; i < s.length; i++) {
			int ct = s[i] == 'a' ? 0 : 1;
			if (ct == currentNode.type)
				currentNode.count++;
			else {
				currentNode = new MinNode();
				currentNode.count = 1;
				currentNode.type = ct;
				N[++nodeCount] = currentNode;
			}
		}
		nodeCount++;
		if (nodeCount < 2)
			return -1;

		Root.initTree(0, nodeCount - 1);
		boolean has2112 = false;

		for (int i = 0; i < nodeCount; i++) {
			if (!has2112 && i < nodeCount - 3) {
				has2112 = ((N[i].count > 1) && (N[i + 1].count == 1) && (N[i + 2].count == 1) && (N[i + 3].count > 1));
			}
			updateMinTo3(i);
			Root.initNode(N[i], i, 0, nodeCount - 1);
		}
		Root.update(0, nodeCount - 1, 0, nodeCount - 1);

		if (Root.totalCount[0] < 3 || Root.totalCount[1] < 3)
			return -1;
		if (Root.max[0] >= 3 && Root.max[1] >= 3)
			return 0;
		if (has2112)
			return 1;
		if (Root.max[0] == 1 && Root.max[1] == 1)
			return 3;
		if (Root.max[0] == 2 && Root.max[1] == 2 && Root.min[0] == 2 && Root.min[1] == 2)
			return 3;
		if (Root.max[0] >= 3)
			return Root.minTo3[1];
		if (Root.max[1] >= 3)
			return Root.minTo3[0];

		int m = BIG_NUMBER;

		for (int i = 0; i < nodeCount && m > 2; i++) {
			MinNode mn = N[i];
			if (mn.count == 2) {
				if (i > 1) {
					m = testSwapNode(i - 2, i, -1, N[i - 1], m);
					m = testSwapNode(i - 2, i, 2, N[i - 1], m);
				}
				if (i < nodeCount - 2) {
					m = testSwapNode(i, i + 2, 1, N[i + 1], m);
					m = testSwapNode(i, i + 2, -2, N[i + 1], m);
				}
			}
		}

		return m;
	}

	int testSwapNode(int base, int target, int testValue, MinNode using, int m) {
		int from = (base == 0 ? 0 : base - 1);
		int to = (target == nodeCount - 1 ? nodeCount - 1 : target + 1);

		N[base].count += testValue;
		N[target].count -= testValue;

		for (int i = from; i <= to; i++)
			updateMinTo3(i);

		Root.update(from, to, 0, nodeCount - 1);

		m = Math.min(m, using.count * Math.abs(testValue) + Root.minTo3[using.type]);

		N[base].count -= testValue;
		N[target].count += testValue;
		for (int i = from; i <= to; i++)
			updateMinTo3(i);

		Root.update(from, to, 0, nodeCount - 1);
		return m;
	}

	private void updateMinTo3(int i) {
		MinNode n = N[i];

		int l = (i > 1) ? N[i - 1].count : BIG_NUMBER;
		int r = (i < nodeCount - 2) ? N[i + 1].count : BIG_NUMBER;

		n.minTo3[n.type] = BIG_NUMBER;
		if (n.count == 1 && i > 1 && i < nodeCount - 2)
			n.minTo3[n.type] = l + r;
		else if (n.count == 2)
			n.minTo3[n.type] = Math.min(l, r);
		else if (n.count > 2)
			n.minTo3[n.type] = 0;
	}
}

Analysis summary

The following issues have been detected: timeout errors.

Analysis

Detected time complexity:

O(N) or O(N**4)

▶

example1
First example test.

✔

▶

example2
Second example test.

✔

▶

example3
Third example test.

✔

▶

example4
Fourth example test.

✔

▶

ans_2
Short test with result -1 or 2. N <= 10.

✔

0.008 s

0.004 s

0.008 s

0.004 s

0.008 s

10.

0.004 s

11.

0.004 s

12.

0.008 s

13.

0.004 s

14.

0.004 s

15.

0.004 s

16.

0.004 s

▶

ans03
Short test with result 0 or 3. N <= 10.

✔

0.004 s

0.008 s

0.004 s

0.008 s

0.004 s

0.008 s

0.012 s

10.

0.008 s

11.

0.004 s

12.

0.004 s

13.

0.008 s

14.

0.004 s

15.

0.004 s

16.

0.004 s

▶

ans14
Short test with result 1 or 4. N <= 16.

✔

0.004 s

10.

0.004 s

11.

0.004 s

12.

0.004 s

13.

0.004 s

14.

0.004 s

15.

0.008 s

16.

0.004 s

17.

0.004 s

18.

0.008 s

▶

ans5
Short test with result 5. N <= 10.

✔

▶

random
Random test cases. N <= 10.

✔

▶

max_answer
Number of required swaps is maximum possible. N <= 8.

✔

▶

medium_big_answer
Answer is bigger than 4. N <= 50. Score x 2.

✔

▶

medium_pattern
S contains repeating pattern. N <= 100. Score x 2.

✔

0.008 s

0.004 s

0.008 s

0.004 s

0.008 s

0.004 s

10.

0.004 s

11.

0.008 s

12.

0.004 s

▶

large_big_answer
Answer is bigger than 4. Score x 2.

✔

▶

large_no_three_consecutive
No three consecutive letters in S are equal. Score x 2.

✘

TIMEOUT ERROR
running time: 4.588 sec., time limit: 0.784 sec.

0.596 s

0.600 s

4.588 s

TIMEOUT ERROR, running time: 4.588 sec., time limit: 0.784 sec.

0.004 s

0.604 s

0.004 s

▶

large_perf
Big performance package. Score x 2.

✘

TIMEOUT ERROR
running time: 0.944 sec., time limit: 0.832 sec.

0.600 s

0.888 s

0.944 s

TIMEOUT ERROR, running time: 0.944 sec., time limit: 0.832 sec.

4.688 s

TIMEOUT ERROR, running time: 4.688 sec., time limit: 0.832 sec.

0.952 s

TIMEOUT ERROR, running time: 0.952 sec., time limit: 0.832 sec.

0.944 s

TIMEOUT ERROR, running time: 0.944 sec., time limit: 0.816 sec.

0.012 s

0.936 s

TIMEOUT ERROR, running time: 0.936 sec., time limit: 0.832 sec.

0.948 s

TIMEOUT ERROR, running time: 0.948 sec., time limit: 0.832 sec.

10.

0.932 s

TIMEOUT ERROR, running time: 0.932 sec., time limit: 0.832 sec.

11.

0.944 s

TIMEOUT ERROR, running time: 0.944 sec., time limit: 0.832 sec.