Added a problem statement

Bit Manipulation/Applications_on_bit.cpp

@@ -0,0 +1,26 @@
+#include <iostream>
+using namespace std;
+
+int getBit(int n,int pos){
+    return((n & (1<<pos))!=0);
+}
+int setBit(int n, int pos){
+    return (n|(1<<pos));
+}
+int clearBit(int n, int pos){
+    int mask=~(1<<pos);
+    return (n & mask);
+}
+int updateBit(int n, int pos, int value){
+    int mask= ~(1<<pos);
+    n = n & mask;
+    return (n| (value<<pos));
+}
+int main()
+{
+    cout<<"get bit is: "<<getBit(5,2)<<endl;
+    cout<<"set bit is: "<<setBit(5,1)<<endl;
+    cout<<"cleared bit is: "<<clearBit(5,2)<<endl;
+    cout<<"updated bit is: "<<updateBit(5,1,1)<<endl;
+return 0;
+}

Problems/SmallestLeft/README.md

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+# Smallest Left
+
+## Problem Statement
+Given an array a of integers of length n, 
+find the nearest smaller number for every element such that the smaller element is on left side.
+If no small element present on the left print -1.
+
+## Algorithm 
+A Simple Solution is to use two nested loops. The outer loop starts from the second element, 
+the inner loop goes to all elements on the left side of the element picked by the outer loop and stops as soon as it finds a smaller element.  
+
+## Time Complexity: 
+O(n2).
+
+## Space Complexity: 
+O(1)

Problems/SmallestLeft/SmallestLeft.cpp

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+#include <bits/stdc++.h>
+using namespace std;
+
+class Solution{
+public:
+    vector<int> leftSmaller(int n, int a[]){
+        // code here
+        vector <int> arr(n,-1);
+        for(int i =1; i<n;i++){
+            int temp = INT_MAX;
+            for(int j=i-1;j>=0;j--){
+                if(a[j]<a[i]){
+                    arr[i]=a[j];
+                    break;
+                }
+            }
+        }
+        return arr;
+    }
+};
+
+int main() {
+    int t;
+    cin>>t;
+    while(t--){
+        int n;
+        cin>>n;
+
+        int a[n];
+        for(int i =0; i<n; i++)
+        cin>>a[i];
+
+        Solution ob;
+        vector<int> ans = ob.leftSmaller(n,a);
+        for(int i =0; i<n; i++)
+        cout<<ans[i]<<" ";
+        cout<<endl;
+    }
+    return 0;
+}