Add Python solution for Codeforces 1538C (Number of Pairs)

- Implemented binary search approach for pair counting
- Time complexity: O(n log n)
- Space complexity: O(n)
- Handles constraints up to n=2×10^5

Author: ishanrajsingh

Problems/Binary-Search/Day-07/sol/Ishan_Raj_Singh/Solution1.py

@@ -0,0 +1,45 @@
+"""
+Approach: Two Pointers approach
+Submission Link: https://codeforces.com/problemset/submission/1538/356144559
+"""
+
+def solve_two_pointers():
+    n, l, r = map(int, input().split())
+    a = list(map(int, input().split()))
+    
+    a.sort()
+    count = 0
+    
+    for i in range(n):
+        min_sum = l - a[i]
+        max_sum = r - a[i]
+        
+        left = i + 1
+        right = n - 1
+        
+        lo, hi = i + 1, n
+        while lo < hi:
+            mid = (lo + hi) // 2
+            if a[mid] < min_sum:
+                lo = mid + 1
+            else:
+                hi = mid
+        left_bound = lo
+        
+        lo, hi = i + 1, n
+        while lo < hi:
+            mid = (lo + hi) // 2
+            if a[mid] <= max_sum:
+                lo = mid + 1
+            else:
+                hi = mid
+        right_bound = lo - 1
+        
+        if left_bound < n and right_bound >= i + 1 and left_bound <= right_bound:
+            count += (right_bound - left_bound + 1)
+    
+    print(count)
+
+t = int(input())
+for _ in range(t):
+    solve_two_pointers()