Alternating Groups III

Alternating Groups III - Problem

There are some red and blue tiles arranged in a circle. You are given an array of integers colors and a 2D array of integers queries.

The color of tile i is represented by colors[i]:

colors[i] == 0 means that tile i is red
colors[i] == 1 means that tile i is blue

An alternating group is a contiguous subset of tiles in the circle with alternating colors (each tile in the group except the first and last one has a different color from its adjacent tiles in the group).

You have to process queries of two types:

queries[i] = [1, size_i]: determine the count of alternating groups with size size_i
queries[i] = [2, index_i, color_i]: change colors[index_i] to color_i

Return an array answer containing the results of the queries of the first type in order.

Note: Since colors represents a circle, the first and the last tiles are considered to be next to each other.

Input & Output

Example 1 — Basic Pattern Counting

$ Input: colors = [0,1,1,0,1], queries = [[1,3],[2,1,0],[1,3]]

› Output: [3,3]

💡 Note: First query [1,3] counts alternating groups of size 3. Groups starting at positions 2: [1,0,1], position 3: [0,1,0], and position 4: [1,0,0] are valid, giving 3 total. After updating index 1 to 0, colors becomes [0,0,1,0,1]. Groups of size 3 starting at positions 1: [0,1,0], position 2: [1,0,1], and position 3: [0,1,0] are valid, giving 3 total.

Example 2 — Size 1 Groups

$ Input: colors = [0,0,1,0], queries = [[1,4],[2,0,1],[1,4]]

› Output: [1,0]

💡 Note: Initially only one alternating group of size 4 exists. After changing colors[0] to 1, no alternating groups of size 4 remain.

Example 3 — All Same Colors

$ Input: colors = [0,0,0], queries = [[1,2]]

› Output: [0]

💡 Note: No alternating groups of size 2 exist when all colors are the same.

Constraints

1 ≤ colors.length ≤ 5 × 10⁴
0 ≤ colors[i] ≤ 1
1 ≤ queries.length ≤ 5 × 10⁴
queries[i][0] ∈ {1, 2}
For type 1 queries: 1 ≤ size_i ≤ colors.length
For type 2 queries: 0 ≤ index_i < colors.length, 0 ≤ color_i ≤ 1

Visualization

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Asked in

G Google 15 f Meta 12 a Amazon 8

The key insight is to efficiently track alternating color patterns in a circular array while handling dynamic updates. The optimal approach uses segment tracking to pre-compute alternating lengths and update only affected regions. Time: O(Q × N), Space: O(N)

Common Approaches

✓ Brute Force

⏱️ Time: O(Q × N × S) Space: O(1)

For type 1 queries, iterate through all possible starting positions and check if each substring of the required size forms an alternating pattern. For type 2 queries, simply update the color at the given index.

Segment Tracking

⏱️ Time: O(Q × N) Space: O(N)

Pre-compute the length of alternating segments starting from each position. For count queries, use these precomputed lengths. For update queries, only recompute segments affected by the color change.

Brute Force — Algorithm Steps

Process each query sequentially
For count queries: check all positions for alternating patterns of given size
For update queries: change the color at specified index

Visualization

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Step-by-Step Walkthrough

Query Processing

Process each query sequentially

Pattern Checking

For count queries, check all positions

Color Updates

For update queries, change color directly

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

static int colors[50000];
static int queries[50000][3];
static int result[50000];

int countAlternatingGroups(int* colors, int n, int size) {
    if (size == 1) return n;
    if (size > n) return 0;
    
    int count = 0;
    for (int start = 0; start < n; start++) {
        int isAlternating = 1;
        for (int i = 1; i < size; i++) {
            int pos = (start + i) % n;
            int prevPos = (start + i - 1) % n;
            if (colors[pos] == colors[prevPos]) {
                isAlternating = 0;
                break;
            }
        }
        if (isAlternating) count++;
    }
    return count;
}

int* solution(int* colors, int colorsSize, int queries[][3], int queriesSize, int* queriesColSize, int* returnSize) {
    int resultIndex = 0;
    
    int colorsCopy[50000];
    for (int i = 0; i < colorsSize; i++) {
        colorsCopy[i] = colors[i];
    }
    
    for (int i = 0; i < queriesSize; i++) {
        if (queries[i][0] == 1) {
            result[resultIndex++] = countAlternatingGroups(colorsCopy, colorsSize, queries[i][1]);
        } else {
            colorsCopy[queries[i][1]] = queries[i][2];
        }
    }
    
    *returnSize = resultIndex;
    return result;
}

int parseColors(char* line, int* colors) {
    int count = 0;
    char* ptr = line + 1; // Skip '['
    
    while (*ptr && *ptr != ']') {
        while (*ptr == ' ' || *ptr == ',') ptr++;
        if (*ptr >= '0' && *ptr <= '9') {
            colors[count++] = strtol(ptr, &ptr, 10);
        } else {
            ptr++;
        }
    }
    return count;
}

int parseQueries(char* line, int queries[][3]) {
    int count = 0;
    char* ptr = line + 2; // Skip "[["
    
    while (*ptr && count < 50000) {
        int querySize = 0;
        
        // Parse one query
        while (*ptr && *ptr != ']' && querySize < 3) {
            while (*ptr == ' ' || *ptr == ',' || *ptr == '[') ptr++;
            if (*ptr >= '0' && *ptr <= '9') {
                queries[count][querySize++] = strtol(ptr, &ptr, 10);
            } else {
                ptr++;
            }
        }
        
        if (querySize > 0) count++;
        
        // Skip to next query
        while (*ptr && *ptr != '[' && *ptr != '\n' && *ptr != '\0') ptr++;
        if (*ptr == '[') ptr++;
        
        if (*ptr == '\n' || *ptr == '\0') break;
    }
    
    return count;
}

int main() {
    char line[100000];
    
    // Read colors
    fgets(line, sizeof(line), stdin);
    int colorsSize = parseColors(line, colors);
    
    // Read queries
    fgets(line, sizeof(line), stdin);
    int dummy[50000];
    int queriesSize = parseQueries(line, queries);
    
    int returnSize;
    int* res = solution(colors, colorsSize, queries, queriesSize, dummy, &returnSize);
    
    printf("[");
    for (int i = 0; i < returnSize; i++) {
        printf("%d", res[i]);
        if (i < returnSize - 1) printf(",");
    }
    printf("]\n");
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(Q × N × S)

Q queries, each count query checks N positions with up to S comparisons

✓ Linear Growth

Space Complexity

O(1)

Only using variables for iteration, no extra data structures

✓ Linear Space

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Input & Output

Constraints

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Related Problems

Common Approaches

Brute Force — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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