diff --git a/Kth Largest Element in an Array.cpp b/Kth Largest Element in an Array.cpp
new file mode 100644
index 0000000..043af1a
--- /dev/null
+++ b/Kth Largest Element in an Array.cpp	
@@ -0,0 +1,10 @@
+class Solution {
+public:
+    int findKthLargest(vector<int>& nums, int k) {
+        
+         int z=nums.size();int c=0,i=0;
+         sort(nums.begin(), nums.end(), greater<int>());
+           int a=nums[k-1];
+            return a;
+        }
+    };
\ No newline at end of file
diff --git a/Plus One.cpp b/Plus One.cpp
new file mode 100644
index 0000000..726b439
--- /dev/null
+++ b/Plus One.cpp	
@@ -0,0 +1,29 @@
+class Solution {
+public:
+    vector<int> plusOne(vector<int>& digits) {
+        
+       int n=digits.size();
+    vector<int>kbc;
+    int c=1;
+    
+    for(int i=n-1; i>=0; i--)
+    {
+        
+        if(digits[i]==9 && c==1) 
+            kbc.push_back(0);
+		
+		
+        else kbc.push_back(digits[i]+c),
+            c=0;
+		
+    }
+    
+	
+    if(c==1) 
+    kbc.push_back(c);
+	
+    reverse(kbc.begin(), kbc.end());
+    return kbc;
+        
+    }
+};
diff --git a/README.md b/README.md
index 06b8eca..b09343f 100644
--- a/README.md
+++ b/README.md
@@ -1,6 +1,27 @@
-We are building a dataset of all Data-Structure and algorithm questions to be accessed later via APIs for our webapp backend.
-
-Please help us contribute.
-
-This dataset would be free to use for all.
-
+Given an integer number n, return the difference between the product of its digits and the sum of its digits.
+
+ 
+
+Example 1:
+
+Input: n = 234
+Output: 15 
+Explanation: 
+Product of digits = 2 * 3 * 4 = 24 
+Sum of digits = 2 + 3 + 4 = 9 
+Result = 24 - 9 = 15
+
+Example 2:
+
+Input: n = 4421
+Output: 21
+Explanation: 
+Product of digits = 4 * 4 * 2 * 1 = 32 
+Sum of digits = 4 + 4 + 2 + 1 = 11 
+Result = 32 - 11 = 21
+
+ 
+
+Constraints:
+
+    1 <= n <= 10^5
diff --git a/README1.md b/README1.md
new file mode 100644
index 0000000..2c4445d
--- /dev/null
+++ b/README1.md
@@ -0,0 +1,29 @@
+Given an array nums. We define a running sum of an array as runningSum[i] = sum(nums[0]…nums[i]).
+
+Return the running sum of nums.
+
+ 
+
+Example 1:
+
+Input: nums = [1,2,3,4]
+Output: [1,3,6,10]
+Explanation: Running sum is obtained as follows: [1, 1+2, 1+2+3, 1+2+3+4].
+
+Example 2:
+
+Input: nums = [1,1,1,1,1]
+Output: [1,2,3,4,5]
+Explanation: Running sum is obtained as follows: [1, 1+1, 1+1+1, 1+1+1+1, 1+1+1+1+1].
+
+Example 3:
+
+Input: nums = [3,1,2,10,1]
+Output: [3,4,6,16,17]
+
+ 
+
+Constraints:
+
+    1 <= nums.length <= 1000
+    -10^6 <= nums[i] <= 10^6
diff --git a/README2.md b/README2.md
new file mode 100644
index 0000000..1d2d1b1
--- /dev/null
+++ b/README2.md
@@ -0,0 +1,25 @@
+Given an integer array nums and an integer k, return the kth largest element in the array.
+
+Note that it is the kth largest element in the sorted order, not the kth distinct element.
+
+You must solve it in O(n) time complexity.
+
+ 
+
+Example 1:
+
+Input: nums = [3,2,1,5,6,4], k = 2
+Output: 5
+
+Example 2:
+
+Input: nums = [3,2,3,1,2,4,5,5,6], k = 4
+Output: 4
+
+ 
+
+Constraints:
+
+    1 <= k <= nums.length <= 105
+    -104 <= nums[i] <= 104
+
diff --git a/Running Sum of 1d Array.cpp b/Running Sum of 1d Array.cpp
new file mode 100644
index 0000000..03d3301
--- /dev/null
+++ b/Running Sum of 1d Array.cpp	
@@ -0,0 +1,16 @@
+class Solution {
+public:
+    vector<int> runningSum(vector<int>& nums) {
+
+        int s=0;
+    
+        for(int i=0;i<nums.size();i++)
+        {
+            
+                s=s+nums[i];
+                   nums[i]=s;            
+        }
+        return nums;
+        
+    }
+};
\ No newline at end of file
diff --git a/Subtract the Product and Sum of Digits of an Integer.cpp b/Subtract the Product and Sum of Digits of an Integer.cpp
new file mode 100644
index 0000000..dbabda4
--- /dev/null
+++ b/Subtract the Product and Sum of Digits of an Integer.cpp	
@@ -0,0 +1,36 @@
+class Solution {
+public:
+    int subtractProductAndSum(int n) {
+        
+        int m,s=1,z,a,s1=0;
+            m=n;
+        a=n;
+        while(m!=0)
+        {
+            int k=m%10;
+            s=s*k;
+            m=m/10;
+            
+        }
+        while(a!=0)
+        {
+            z=a%10;
+            s1=s1+z;
+            a=a/10;
+        }
+        int res=s-s1;
+       return res;
+        
+    }
+    int main()
+    {
+        int n;
+        cin>>n;
+       int a= subtractProductAndSum(n);
+        cout<<a<<endl;
+        return 0;
+        
+        
+        
+    }
+};
\ No newline at end of file