diff --git a/Domains/CompetitiveProgramming/Programs/C++/PoliceRecruits.cpp b/Domains/CompetitiveProgramming/Programs/C++/PoliceRecruits.cpp
new file mode 100644
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+++ b/Domains/CompetitiveProgramming/Programs/C++/PoliceRecruits.cpp
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+/*
+Problem: Police Recruits
+Platform: Codeforces
+Problem Code: 427A
+Difficulty: Easy
+Link: https://codeforces.com/problemset/problem/427/A
+
+Problem Statement:
+Given a sequence of events representing crimes and police recruitments:
+- Each -1 represents a crime.
+- Each positive number represents the number of police officers recruited.
+When a crime occurs and there are no available officers, it goes untreated. 
+Find the total number of untreated crimes.
+
+Example:
+Input:
+    7
+    -1 -1 1 1 -1 -1 1
+
+Output:
+    2
+
+Approach:
+1. Traverse through the list of events.
+2. Maintain a count of active police officers.
+3. If a crime occurs and no officer is available, increase the untreated crimes count.
+4. Otherwise, assign one available officer to handle the crime.
+
+Time Complexity: O(n)
+Space Complexity: O(1)
+
+Contributor: Paila-Sahitya
+
+*/
+
+#include<bits/stdc++.h>
+using namespace std;
+ 
+int policeRecruits(int n, vector<int> &arr){
+    int crimeCount=0;
+    int activePolice=0;
+    for(int i=0;i<n;i++){
+        if(arr[i]==-1 && activePolice==0) crimeCount++;
+        else if(arr[i]==-1){
+            activePolice--;
+        }
+        else activePolice+=arr[i];
+    }
+    return crimeCount;
+}
+ 
+int main(){
+    int n;
+    cin>>n;
+    vector<int> arr(n);
+    for(int i=0;i<n;i++){
+        cin>>arr[i];
+    }
+    cout<<policeRecruits(n, arr)<<endl;
+    return 0;
+}
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diff --git a/Domains/CompetitiveProgramming/Programs/C++/Twins.cpp b/Domains/CompetitiveProgramming/Programs/C++/Twins.cpp
new file mode 100644
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+++ b/Domains/CompetitiveProgramming/Programs/C++/Twins.cpp
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+/*
+Problem: Twins
+Platform: Codeforces
+Problem Code: 160A
+Difficulty: Easy
+Link: https://codeforces.com/problemset/problem/160/A
+
+Problem Statement: 
+Two brothers have a collection of coins of different values.
+The goal is to find the minimum number of coins one brother must take 
+so that the total value of his coins becomes strictly greater than the 
+total value of the remaining coins.
+
+Example:
+Input:
+    4
+    3 3 4 3
+Output:
+    2
+
+Approach: 
+1. Sort the coins in descending order.
+2. Keep picking the largest coins until the sum of selected coins > sum of remaining coins.
+3. Return the number of coins picked.
+
+Time Complexity: O(n log n)
+Space Complexity: O(1)
+
+Contributor: Paila-Sahitya
+*/
+
+#include<bits/stdc++.h>
+using namespace std;
+ 
+int minimumCoins(int n, vector<int>&arr){
+    sort(arr.begin(), arr.end(), greater<int>());
+    int total=0;
+    for(int i=0;i<n;i++){
+        total+=arr[i];
+    }
+    int sum=0;
+    int count=0;
+    for(int i=0;i<n;i++){
+        sum+=arr[i];
+        count++;
+        if(sum>total-sum) return count;
+    }
+    return count;
+}
+
+int main(){
+    int n;
+    cin>>n;
+    vector<int> arr(n);
+    for(int i=0;i<n;i++){
+        cin>>arr[i];
+    }
+    cout<<minimumCoins(n, arr)<<endl;
+    return 0;
+}
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