Arithmetic Slices II Subsequence LeetCode Solution

Here, We see Arithmetic Slices II Subsequence LeetCode Solution. This Leetcode problem is done in many programming languages like C++, Java, JavaScript, Python, etc. with different approaches.

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Arithmetic Slices II Subsequence LeetCode Solution

Arithmetic Slices II Subsequence LeetCode Solution

Problem Statement

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.

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 32-bit integer.

Example 1:
Input: nums = [2,4,6,8,10]
Output: 7
Explanation: All arithmetic subsequence slices are: [2,4,6] [4,6,8] [6,8,10] [2,4,6,8] [4,6,8,10] [2,4,6,8,10] [2,6,10]

Example 2:
Input: nums = [7,7,7,7,7]
Output: 16
Explanation: Any subsequence of this array is arithmetic.

Arithmetic Slices II Subsequence LeetCode Solution C++

class Solution {
public:
    int numberOfArithmeticSlices(vector<int>& nums) {
        int n = nums.size();
        int total_count = 0;
        std::vector<std::unordered_map<int, int>> dp(n);
        for (int i = 1; i < n; ++i) {
            for (int j = 0; j < i; ++j) {
                long long diff = static_cast<long long>(nums[i]) - nums[j];
                if (diff > INT_MAX || diff < INT_MIN)
                    continue; 
                int diff_int = static_cast<int>(diff);
                dp[i][diff_int] += 1; 
                if (dp[j].count(diff_int)) {
                    dp[i][diff_int] += dp[j][diff_int];
                    total_count += dp[j][diff_int];
                }
            }
        }
        return total_count;        
    }
};Code language: PHP (php)

Arithmetic Slices II Subsequence LeetCode Solution Java

class Solution {
    public int numberOfArithmeticSlices(int[] nums) {
        int n = nums.length;
        int total_count = 0;
        HashMap<Integer, Integer>[] dp = new HashMap[n];
        for (int i = 0; i < n; ++i) {
            dp[i] = new HashMap<>();
        }
        for (int i = 1; i < n; ++i) {
            for (int j = 0; j < i; ++j) {
                long diff = (long) nums[i] - nums[j]; 
                if (diff > Integer.MAX_VALUE || diff < Integer.MIN_VALUE) {
                    continue; 
                }
                int diffInt = (int) diff;
                dp[i].put(diffInt, dp[i].getOrDefault(diffInt, 0) + 1);  
                if (dp[j].containsKey(diffInt)) {
                    dp[i].put(diffInt, dp[i].get(diffInt) + dp[j].get(diffInt));
                    total_count += dp[j].get(diffInt);
                }
            }
        }
        return total_count;
    }
}Code language: PHP (php)

Arithmetic Slices II Subsequence LeetCode Solution JavaScript

var numberOfArithmeticSlices = function(nums) {
    const n = nums.length;
    let total_count = 0;
    const dp = new Array(n).fill().map(() => new Map());
    for (let i = 1; i < n; ++i) {
        for (let j = 0; j < i; ++j) {
            const diff = nums[i] - nums[j];
            if (dp[j].has(diff)) {
                dp[i].set(diff, (dp[i].get(diff) || 0) + dp[j].get(diff));
                total_count += dp[j].get(diff);
            }
            dp[i].set(diff, (dp[i].get(diff) || 0) + 1);
        }
    }
    return total_count;
};Code language: JavaScript (javascript)

Arithmetic Slices II Subsequence LeetCode Solution Python

class Solution(object):
    def numberOfArithmeticSlices(self, nums):
        n = len(nums)
        dp = [[0] * n for _ in range(n)]
        map = {}
        for i in range(n):
            temp = nums[i]
            if temp not in map:
                map[temp] = []
            map[temp].append(i)

        total_sum = 0
        for i in range(1, n):
            for j in range(i + 1, n):
                a = 2 * nums[i] - nums[j]
                if a in map:
                    for k in map[a]:
                        if k < i:
                            dp[i][j] += dp[k][i] + 1
                        else:
                            break
                total_sum += dp[i][j]
        return total_sumCode language: HTML, XML (xml)
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