Simple approach that beats 100% without dp
Intuition\n Describe your first thoughts on how to solve this problem. \nThe intuition behind the code is to calculate the minimum number of streets that need t
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe intuition behind the code is to calculate the minimum number of streets that need to be traveled to reach one house from another in a city with a specific layout. The city is represented as a cycle of houses with two tails. The cycle is formed by the houses from houseX to houseY, and the tails are the houses before houseX and after houseY.\n\nThe code calculates the number of pairs of houses for each possible distance in three parts:\n\nPairs within the Cycle: The code first calculates the number of pairs of houses within the cycle. For each distance less than half of the cycle, there are twice the number of pairs as there are houses in the cycle. If the cycle\u2019s length is even, there are as many pairs at a distance of half the cycle as there are houses in the cycle.\n\nTail to Tail Pairs: The code then calculates the number of pairs of houses that are both in the tails. For each distance from 1 to the total length of the tails, there are twice the number of pairs as there are houses remaining in the tails. If houseX is not equal to houseY, it subtracts 2 from the number of pairs at a distance of 0 and for each tail, it subtracts 2 from the number of pairs at each distance from 1 to the length of the tail.\n\nPairs from Tails to the Cycle: Finally, the code calculates the number of pairs of houses where one house is in a tail and the other house is in the cycle. For each tail and each distance from 1 to the sum of the tail and half the cycle, it adds four times the minimum of the tail, half the cycle, the distance, and the sum of the tail and half the cycle minus the distance to the number of pairs at that distance. If the cycle\u2019s length is even, it also adds 2 to the number of pairs at each distance from half the cycle to the sum of half the cycle and the tail.\n# Approach\n<!-- Describe your approach to solving the problem. -->\nThe city is represented as a cycle of houses with two tails. The cycle is formed by the houses from houseX to houseY, and the tails are the houses before houseX and after houseY.\n\nInitialize Variables: The function first checks if houseY is less than houseX and if so, swaps them. It then initializes an array result of size totalHouses with all elements set to 0. It also calculates the cycle, halfCycle, tail1, tail2, and totalTailToTail.\n\nCalculate Pairs within the Cycle: The function then calculates the number of pairs of houses within the cycle (the direct path between houseX and houseY). For each distance i less than halfCycle, there are cycle * 2 pairs of houses. If cycle is even, there are cycle pairs of houses at a distance of halfCycle.\n\nCalculate Tail to Tail Pairs: The function then calculates the number of pairs of houses that are both in the tails (the houses before houseX and after houseY). For each distance i from 1 to totalTailToTail, there are (totalTailToTail - i) * 2 pairs of houses. If houseX is not equal to houseY, it subtracts 2 from the number of pairs at a distance of 0 and for each tail, it subtracts 2 from the number of pairs at each distance from 1 to the length of the tail.\n\nCalculate Pairs from Tails to the Cycle: Finally, the function calculates the number of pairs of houses where one house is in a tail and the other house is in the cycle. For each tail and each distance i from 1 to tail + halfCycle, it adds 4 * min([tail, halfCycle, i, tail + halfCycle - i]) to the number of pairs at distance i. If cycle is even, it also adds 2 to the number of pairs at each distance from halfCycle to halfCycle + tail.\n\nReturn Result: The function returns the result array, which contains the total number of pairs of houses for each distance from 0 to n - 1.\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\nO(n)\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\nO(n)\n# Code\n```\nclass Solution:\n    def countOfPairs(self, n: int, x: int, y: int) -> List[int]:\n        houseX=x\n        houseY=y\n        totalHouses=n\n        if houseY < houseX:\n            houseX, houseY = houseY, houseX\n\n        result = [0] * totalHouses\n        cycle = houseY - houseX + 1\n        halfCycle = (cycle - 1) >> 1\n        tail1 = houseX - 1\n        tail2 = totalHouses - houseY\n        totalTailToTail = tail1 + tail2 + 1 + (houseX != houseY)\n\n        for i in range(halfCycle):\n            result[i] = cycle * 2\n        if cycle % 2 == 0:\n            result[halfCycle] = cycle\n            \n        for i in range(1, totalTailToTail):\n            result[i - 1] += (totalTailToTail - i) * 2\n        if houseX != houseY:\n            result[0] -= 2\n            for tail in [tail1, tail2]:\n                for i in range(1, tail + 1):\n                    result[i] -= 2\n\n        for tail in [tail1, tail2]:\n            for i in range(1, tail + halfCycle):\n                result[i] += 4 * min([tail, halfCycle, i, tail + halfCycle - i])\n            if cycle % 2 == 0:\n                for i in range(1, tail + 1):\n                    result[i + halfCycle] += 2\n\n        return result\n\n            \n```
['Python3']