Effortlessly solve complex problems with this innovative tool
To understand problem easily use below tool:\nhttps://drawtocode.onrender.com/problems/find-the-number-of-winning-players ( paste this url in your tab)\n\n\n\n#
# To understand problem easily use below tool:\nhttps://drawtocode.onrender.com/problems/find-the-number-of-winning-players ( paste this url in your tab)\n![drawtocode.png](https://assets.leetcode.com/users/images/90afd410-3a6f-4110-b0e4-9a8809d114a4_1722506314.2638276.png)\n\n\n# Intuition\nProblem Statement:\n\nWe are given a 2D array with n rows and 2 columns. The first column represents the ball number, and the second column represents the color of the ball. The task is to return the count of balls that satisfy the following condition:\n\nFor a particular ball number in the first column, it should be less than the count of balls with the same color as indicated by the second column.\n\nApproach:\n\nSorting the Array:\n\nFirst, sort the array based on the first column (ball number) in ascending order.\nWithout altering the first column, sort the array again based on the second column (ball color) in ascending order.\nInitial Setup:\n\nThere is a base condition for ball number 0. If this ball number is present in the array, it automatically satisfies the condition. In this case, increment the count by one. Additionally, set prev1 to 0 (representing the ball number) and prev2 to the color of the ball in the last occurrence of ball number 0.\nIncrement the pointer so that it points to the ball number next to 0.\nIterative Process:\n\nIf the base condition is not met, point to the second row with a pointer ti, and assign the current count of balls with the same color as the ti-th ball number to ccount. Set prev1 and prev2 to the first and second column values of the 0th row, respectively.\nIterate through the array using this ti variable.\nSkipping Unnecessary Iterations:\n\nWithin the loop, there is a condition that allows skipping the remaining part of the code:\ncpp\nCopy code\nif (pick[i][0] - ccount > (pick.size() - i) - 1) {\n    continue;\n}\nThis condition checks if the current ball number is greater than the number of remaining rows in the array. If true, skip the further part of the loop.\nUpdating the Count:\n\nIf the current ball number (pick[i][0]) and ball color (pick[i][1]) match prev1 and prev2, respectively, increment the current count (ccount).\nThere is another condition to check:\ncpp\nCopy code\nif (ccount == pick[i][1] + 1) {\n    count++;\n}\nThis condition ensures that the count is incremented if the current count of balls (ccount) matches the color of the current ball plus one.\nResetting the Counters:\n\nIf the current ball number or color does not match prev1 or prev2, update prev1 to the current ball number, prev2 to the current ball color, and reset ccount to 1.\nFinal Step:\n\nAfter iterating through the entire loop, return the final count.\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\nO(n)\n- Space complexity:\nO(1)\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```cpp []\nclass Solution {\npublic:\nstatic bool sortByFirstAnd2ndElement(const vector<int>& a, const vector<int>& b) {\n     if (a[0] == b[0])\n            return a[1] < b[1];\n        return a[0] < b[0];\n}\n\n    int winningPlayerCount(int n, vector<vector<int>>& pick) {\n      \n        sort(pick.begin(),pick.end(),sortByFirstAnd2ndElement);\n       \n\n\n  \n        int count = 0;\n        int ccount = 1;\n        int ti =0;\n        int prev1 = pick[0][0];\n        int prev2 = pick[0][1];\n        while(ti < pick.size() && pick[ti][0] == 0 )\n        {\n            count =1;\n             prev1 = pick[ti][0];\n         prev2 = pick[ti][1];\n            ti++;\n        }\n \nif(ti == 0){\n    ti =1;\n}\n       for(int i = ti; i < pick.size() ;i++)\n       {\n\n        if(pick[i][0] - ccount > (pick.size() - i)-1){\n            continue;\n        }\n        if(prev1 == pick[i][0] && prev2 == pick[i][1]){\n            ccount++;\n           \n            if(ccount == pick[i][0]+1){\n              \n                count++;\n            }\n\n        }else{\n            prev1 = pick[i][0];\n            prev2 = pick[i][1];\n            ccount =1;\n            \n        }\n\n       }\n        return count;\n    }\n};\n```
['Sorting', 'C++']