Beats 100% || Simplest explanation
Intuition\nlet no be large and small \n4large  + 2small = tomato .... eqn(i)\nlarge + small = cheese ..... eqn(ii)\nfrom both equation \nlarge = (tomato)/2 - ch
# Intuition\nlet no be large and small \n4*large  + 2*small = tomato .... eqn(i)\nlarge + small = cheese ..... eqn(ii)\nfrom both equation \nlarge = (tomato)/2 - cheese \nsmall = cheese - large \n\n.\n.\nPls upvote if you liked !!\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution {\npublic:\n    vector<int> numOfBurgers(int tomato, int cheese) \n    {\n        \n        if(tomato % 2!=0 || cheese*2> tomato || cheese >tomato)\n        return { } ;\n        int large  = tomato/2 - cheese ; \n        int small = cheese - large ; \n        if(large >=0 && small>=0)\n        return {large,small};\n        else return { };\n\n\n\n    }\n};\n```
['C++']