Merge pull request #509 from ishanrajsingh/day9-q1

Add salary queries solution

Author: Dheeraj-06

Problems/Data-structures/Day-09/sol/ishanrajsingh/Solution1.py

@@ -0,0 +1,55 @@
+#Salary Queries - A company has n employees with certain salaries. We have to keep track of the salaries and process queries.
+#Approach - Use coordinate compression + Binary Indexed Tree
+#Time Complexity - O((n+q)log(n+q))
+
+import sys
+input = sys.stdin.readline
+
+n, q = map(int, input().split())
+p = list(map(int, input().split()))
+
+queries = []
+vals = set(p)
+
+for _ in range(q):
+    s = input().split()
+    if s[0] == '!':
+        k, x = int(s[1]), int(s[2])
+        queries.append(('!', k - 1, x))
+        vals.add(x)
+    else:
+        a, b = int(s[1]), int(s[2])
+        queries.append(('?', a, b))
+        vals.add(a)
+        vals.add(b)
+
+vals = sorted(vals)
+idx = {v: i + 1 for i, v in enumerate(vals)}
+
+m = len(vals)
+bit = [0] * (m + 1)
+
+def add(i, v):
+    while i <= m:
+        bit[i] += v
+        i += i & -i
+
+def sum_(i):
+    s = 0
+    while i > 0:
+        s += bit[i]
+        i -= i & -i
+    return s
+
+for x in p:
+    add(idx[x], 1)
+
+for qu in queries:
+    if qu[0] == '!':
+        i, x = qu[1], qu[2]
+        add(idx[p[i]], -1)
+        p[i] = x
+        add(idx[x], 1)
+    else:
+        a, b = qu[1], qu[2]
+        print(sum_(idx[b]) - sum_(idx[a] - 1))