diff --git a/C/Algorithms/Dynamic-Programming/Longest_Common_Subsequence.c b/C/Algorithms/Dynamic-Programming/Longest_Common_Subsequence.c
new file mode 100644
index 00000000..cc540614
--- /dev/null
+++ b/C/Algorithms/Dynamic-Programming/Longest_Common_Subsequence.c
@@ -0,0 +1,47 @@
+#include <stdio.h>
+#include <stdlib.h>
+#include <string.h>
+
+int main()
+{
+    int l1,l2,i,j,check=0;
+    char *ar1=(char*)malloc(50*sizeof(char));
+    char *ar2=(char*)malloc(50*sizeof(char));
+    printf("enter string 1 and string 2\n");
+    scanf("%s %s",ar1,ar2);
+    l1=strlen(ar1);
+    l2=strlen(ar2);
+
+    //CREATING DP TABLE
+    int **arr=(int **)malloc((l1+1)*sizeof(int *));//2D array with constitutes the DP table
+    for(i=0;i<=l1;i++)
+        arr[i]=(int *)malloc((l2+1)*sizeof(int *));
+
+    //PREPROCESSING DP TABLE
+    for(i=0;i<=l1;i++)
+        arr[i][0]=0;
+    for(i=0;i<=l2;i++)
+        arr[0][i]=0;
+
+    //DP ITERATION STARTS
+    for(i=1;i<=l1;i++)
+    {
+        for(j=1;j<=l2;j++)
+        {
+            if(ar1[i-1]==ar2[j-1])
+                arr[i][j]=arr[i-1][j-1]+1;
+            else
+            {
+                if(arr[i-1][j]>arr[i][j-1])
+                    arr[i][j]=arr[i-1][j];
+                else
+                    arr[i][j]=arr[i][j-1];
+            }
+        }
+    }
+    //arr[l1][l2] contains the length of the longest common subsequence
+    printf("The length of Longest Common Subsequence is %d\n", arr[l1][l2]);
+    
+    return 0;
+}
+
diff --git a/C/Algorithms/Dynamic-Programming/readme.md b/C/Algorithms/Dynamic-Programming/readme.md
index b4e1e362..2f5388d2 100644
--- a/C/Algorithms/Dynamic-Programming/readme.md
+++ b/C/Algorithms/Dynamic-Programming/readme.md
@@ -1 +1,6 @@
 ### Dynamic-Programming
+- [Longest Common Subsequence](Longest_Common_Subsequence.c)
+
+    Given two sequences, find the length of longest subsequence present in both of them.
+
+    Time complexity = O(mn) where "m" is the length of the 1st sequence and "n" is the length of the 2nd sequence