Leetcode Arithmetic Slices II – Subsequence problem solution YASH PAL, 31 July 2024 In this Leetcode Arithmetic Slices II – Subsequence problem solution you have given an integer array nums, return the number of all the arithmetic subsequences of nums.A sequence of numbers is called arithmetic if it consists of at least three elements and if the difference between any two consecutive elements is the same.For example, [1, 3, 5, 7, 9], [7, 7, 7, 7], and [3, -1, -5, -9] are arithmetic sequences.For example, [1, 1, 2, 5, 7] is not an arithmetic sequence.A subsequence of an array is a sequence that can be formed by removing some elements (possibly none) of the array. For example, [2,5,10] is a subsequence of [1,2,1,2,4,1,5,10].The test cases are generated so that the answer fits in a 32-bit integer. Problem solution in Python.class Solution: def numberOfArithmeticSlices(self, A: List[int]) -> int: dic={} for i in range(len(A)): dic[i]={} ans=0 for i in range(len(A)): for j in range(i): d=A[i]-A[j] if d not in dic[j]: if d in dic[i]: dic[i][d]+=1 else: dic[i][d]=1 else: if d not in dic[i]: dic[i][d]=dic[j][d]+1 ans+=dic[j][d] else: dic[i][d]+=dic[j][d]+1 ans+=dic[j][d] return ans Problem solution in Java.public int numberOfArithmeticSlices(int[] A) { int l = A.length; if(l <3){ return 0; int[] left = new int[l]; int[] right = new int[l]; left[0] = 0; right[l-1] = 0; for(int i=1;i<l;i++){ left[i] = A[i] - A[i-1]; } for(int i=l-2;i>=0;i--){ right[i] = A[i+1] - A[i]; } int prev = 0; int[] result = new int[l]; for(int i=1;i<l-1;i++){ if(right[i] == left[i]){ result[i] = 1; if(right[i-1] == left[i-1]){ result[i] = result[i]+result[i-1]; } } } int sum=0; for(int i=0;i<l;i++){ sum+=result[i]; } return sum; } Problem solution in C++.class Solution { public: int numberOfArithmeticSlices(vector<int>& nums) { int ans = 0, n = size(nums); vector<unordered_map<int, int>> freq(n); for (size_t i = 0; i < n; ++i) { for (size_t ii = 0; ii < i; ++ii) { long temp = (long) nums[i] - nums[ii]; if (temp > INT_MAX || temp < INT_MIN) continue; int diff = nums[i] - nums[ii]; ans += freq[ii][diff]; freq[i][diff] += 1 + freq[ii][diff]; } } return ans; } }; coding problems solutions