[Lemo Puzzle] Two array, boxes and lemos stand for size of boxes and size of lemos. Every lemo can be put into a box which has equal or bigger size. Pick a start point i from lemos array, find a possible box and repeat this action for i+1, i+2. All lemos in boxes should keep their original orders. What is the longest sequence of lemos in boxes. …

main.cpp

#include <iostream>

int boxes[10] = {5, 3, 2, 7, 10, 9, 1, 6, 4, 8};
int lemons[10] = {2, 6, 7, 1, 10, 5, 4, 3, 9, 8};

int main() {
    int globalMax = -1;
    
    for (int k = 0; k < 10; k++) {
        int i = k, j = 0;
        std::cout << "Current: " << i << std::endl;

        for (int m = 0; m < 10; m++) std::cout << boxes[m] << ' ';
        std::cout << std::endl;

        while (i < 10 && j < 10) {
            if (lemons[i] <= boxes[j]) {
                std::cout << lemons[i] << ' ';
                i++;
            } else {
                std::cout << "  ";
            }
            j++;
        }

        int localMax = i - k;
        globalMax = globalMax < localMax ? localMax : globalMax;

        std::cout << std::endl << "--> Max length: " << localMax;
        std::cout << std::endl << std::endl;
    }

    std::cout << std::endl << "--> Global Max: " << globalMax << std::endl;
}