Added Sanad's Solution, May 6th

05- May/06- Number of Subsequences That Satisfy the Given Sum Condition/06- Number of Subsequences That Satisfy the Given Sum Condition (Omar Sanad).cpp

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+// author : Omar Sanad
+
+// In this problem we can say that at each idx what is the number of valid subsequences,
+// that this nums[idx] is the smallest number in it
+class Solution {
+public:
+
+    // define some macros for easy use
+    #define ll long long
+    #define MOD int(1e9 + 7)
+
+    // A binary power function that works in O(log(POWER))
+    // We implemented this function to calculate the power and taking the mode
+    ll fastpower(ll base, ll po, ll md) {
+        ll ANS = 1;
+        while (po){
+            if (po & 1)
+                ANS = (ANS * base) % md;
+            base *= base, base %= md;
+            po >>= 1;
+        }
+        return ANS;
+    }
+
+
+    int numSubseq(vector<int>& nums, int target) {
+
+        // we sort the vector
+        sort(nums.begin(), nums.end());
+
+        // declare a variable to store the answer
+        int Answer = 0;
+
+        // iterate over all the elements
+        for (int i = 0, idx; i < nums.size(); i++) {
+
+            // to avoid time limit in the binary search 
+            // we have to search for an element that is bigger than me
+            if (nums[i] > target - nums[i])
+                continue;
+
+            // get the maximum element that meet the codition with nums[i]
+            // where nums[i] + nums[idx] <= target
+            idx = upper_bound(nums.begin() + i, nums.end(), target - nums[i]) - nums.begin();
+            idx--;
+
+            // then we count the number of subsequences that start from this number
+            // every element except v[i]
+            // we two possiblities for it, either take or leave
+            Answer = (Answer + fastpower(2, idx - i, MOD)) % MOD;
+        }
+
+        // return the answer to the problem
+        return Answer;
+    }
+};