1. Sort Colors

Given a array nums containing elements representing red, white, and blue colors (denoted by integers 0, 1, and 2 respectively), sort the array in-place so that objects of the same color are adjacent and arranged in the order red, white, and blue.

The solution can be implemented using bubble sort:

void sortColors(int* nums, int numsSize) {
    for (int i = 0; i < numsSize; i++) {
        for (int j = 0; j < numsSize - i - 1; j++) {
            if (nums[j] > nums[j + 1]) {
                int temp = nums[j];
                nums[j] = nums[j + 1];
                nums[j + 1] = temp;
            }
        }
    }
}

2. Search in 2D Matrix

Given a matrix where each row is sorted in non-decreasing order and the first element of each row is greater than the last element of the previous row, determine whether a given target value exists in the matrix.

A straightforwrad approach involves traversing all elements:

bool searchMatrix(int** matrix, int matrixSize, int* matrixColSize, int target) {
    for (int i = 0; i < matrixSize; i++) {
        for (int j = 0; j < *matrixColSize; j++) {
            if (matrix[i][j] == target)
                return true;
        }
    }
    return false;
}

Alternatively, binary search can be used by treating the 2D matrix as a 1D array. The mapping from 1D index mid to 2D coordinates is defined as follows:

• Row index: row = mid / totalColumns
• Column index: col = mid % totalColumns

Here's the implementation using binary search:

bool searchMatrix(int** matrix, int matrixSize, int* matrixColSize, int target) {
    int left = 0;
    int right = (*matrixColSize) * matrixSize - 1;
    int mid;
    
    while (left <= right) {
        mid = left + (right - left + 1) / 2;
        int row = mid / (*matrixColSize);
        int col = mid % (*matrixColSize);
        
        if (matrix[row][col] == target)
            return true;
        
        if (left >= right)
            break;
        
        if (matrix[row][col] < target)
            left = mid;
        else
            right = mid - 1;
    }
    return false;
}