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+# [FrogRiverOne](https://codility.com/programmers/lessons/4-counting_elements/)
+Find the earliest time when a frog can jump to the other side of a river.
+
+### Solution (JavaScript)
+Similar to the previous problem, our approach is to loop through A and everytime we encounter a position store it in a second array along with the time, effectively reversing the array. We can then loop through the second array and find the maximum time value before we get to X
+    
+__[Test Score: 100%](https://codility.com/demo/results/training3D3DBZ-46Q/)__
+
+```js
+function solution(X, A) {
+    let B = []
+    
+    for (let i = 0; i < A.length; ++i) {
+        if ( !B[A[i]] ) B[A[i]] = i
+    }
+    
+    let max_time = -1
+    for (let i = 1; i <= X; ++i) {
+        if ( typeof B[i] === "undefined" ) return -1
+        if ( B[i] > max_time ) max_time = B[i] 
+    }
+    
+    return max_time
+}
+```