diff --git a/LeetCode/Algorithms/Hard/CountTheNumberOfPowerfulIntegers.cpp b/LeetCode/Algorithms/Hard/CountTheNumberOfPowerfulIntegers.cpp
new file mode 100644
index 0000000..d65b1de
--- /dev/null
+++ b/LeetCode/Algorithms/Hard/CountTheNumberOfPowerfulIntegers.cpp
@@ -0,0 +1,67 @@
+class Solution {
+    public:
+        vector<int> powerfulIntegers(int x, int y, int bound) {
+
+        }
+};
+
+
+/*
+
+16 digit 
+
+10^15 >> 10^8 
+
+pow(10^15, 0.5) == logN ~ 10^8
+
+Observations 
+
+1. Start < Finish,  example: s = 1500, finish = 1234 
+
+s may not be < finish 
+
+Start can be either <= or > 's' 
+
+2. s[i] <= limit 
+
+But this is not necessarily true for start & finish 
+
+example start = 142 
+finish = 24638
+limit = 4 
+s = 214 
+
+3. max no <= 16 digits 
+finish is max No 
+finish <= 10^15 
+
+Idea
+
+start = 15 
+finish = 2134 
+limit = 7 
+s = 10 
+
+Goal: find all nos from start to finish with 's' as suffix 
+ex: 110, 210, 2110, 410, etc 
+
+{count start -> finish} = count (1 -> finish) - count (1 -> start) 
+= x - y 
+
+count(15, 2134) = count(1, 2134) - count(1, 14)  = 18 - 1 
+ = 17 
+
+Boundary Cases: 
+
+If start < s
+then count(start, finish) = count(1, finish) 
+
+example start = 8, s = "10" 
+
+If finish < s then total count = 0 always 
+
+Generating numbers: Digit-wise 
+
+
+
+*/
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