# Arithmetic Slices II - Subsequence Problem

## Description

LeetCode Problem 446.

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 32-bit integer.

Example 1:

``````1
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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:

``````1
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Input: nums = [7,7,7,7,7]
Output: 16
Explanation: Any subsequence of this array is arithmetic.
``````

Constraints:

• 1 <= nums.length <= 1000
• -2^31 <= nums[i] <= 2^31 - 1

## Sample C++ Code

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class Solution {
public:
int numberOfArithmeticSlices(vector<int>& nums) {
vector<unordered_map<long long int,long long int>> dp(nums.size());
long long int ans = 0;

for (int i = 1; i < nums.size(); i ++) {
long long int currele = nums[i];

for (int j = i - 1; j >= 0; j --) {
long long int diff = currele - nums[j];
if (dp[j].find(diff) == dp[j].end()) {
ans += 0;
dp[i][diff] = dp[i][diff] + 1;
}
else {
ans += dp[j][diff];
dp[i][diff] = dp[i][diff] + 1 + dp[j][diff];
}
}
}
return ans;
}
};
``````