Created
March 10, 2018 18:13
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JS Bin // source https://jsbin.com/tihube
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| <!DOCTYPE html> | |
| <html> | |
| <head> | |
| <meta charset="utf-8"> | |
| <meta name="viewport" content="width=device-width"> | |
| <title>JS Bin</title> | |
| </head> | |
| <body> | |
| <script id="jsbin-javascript"> | |
| //check if the index from A[0]=1 to A[K] = X are covered | |
| // return K | |
| /* | |
| 1. FrogRiverOne | |
| Find the earliest time when a frog can jump to the other side of a river. | |
| */ | |
| function solution(X, A) { | |
| let a = A.length; | |
| let set = new Set(A); // remove duplicates | |
| let t =-1, K=0; | |
| if(set.size < X){ | |
| return -1; | |
| } | |
| for(let i=1; i<=X ; i++){ | |
| K = A.indexOf(i); | |
| //console.log(i +", "+ K); | |
| if(K >= 0 ){ | |
| t = K > t ? K : t; | |
| }else{ | |
| return -1; | |
| } | |
| } | |
| return t; | |
| } | |
| let X = 5, A = []; | |
| A[0] = 1 | |
| A[1] = 3 | |
| A[2] = 1 | |
| A[3] = 4 | |
| A[4] = 2 | |
| A[5] = 3 | |
| A[6] = 5 | |
| A[7] = 4 | |
| console.log(solution(X, A)) ; // 6 | |
| </script> | |
| <script id="jsbin-source-javascript" type="text/javascript"> //check if the index from A[0]=1 to A[K] = X are covered | |
| // return K | |
| /* | |
| 1. FrogRiverOne | |
| Find the earliest time when a frog can jump to the other side of a river. | |
| */ | |
| function solution(X, A) { | |
| let a = A.length; | |
| let set = new Set(A); // remove duplicates | |
| let t =-1, K=0; | |
| if(set.size < X){ | |
| return -1; | |
| } | |
| for(let i=1; i<=X ; i++){ | |
| K = A.indexOf(i); | |
| //console.log(i +", "+ K); | |
| if(K >= 0 ){ | |
| t = K > t ? K : t; | |
| }else{ | |
| return -1; | |
| } | |
| } | |
| return t; | |
| } | |
| let X = 5, A = []; | |
| A[0] = 1 | |
| A[1] = 3 | |
| A[2] = 1 | |
| A[3] = 4 | |
| A[4] = 2 | |
| A[5] = 3 | |
| A[6] = 5 | |
| A[7] = 4 | |
| console.log(solution(X, A)) ; // 6</script></body> | |
| </html> |
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Learn more about bidirectional Unicode characters
| //check if the index from A[0]=1 to A[K] = X are covered | |
| // return K | |
| /* | |
| 1. FrogRiverOne | |
| Find the earliest time when a frog can jump to the other side of a river. | |
| */ | |
| function solution(X, A) { | |
| let a = A.length; | |
| let set = new Set(A); // remove duplicates | |
| let t =-1, K=0; | |
| if(set.size < X){ | |
| return -1; | |
| } | |
| for(let i=1; i<=X ; i++){ | |
| K = A.indexOf(i); | |
| //console.log(i +", "+ K); | |
| if(K >= 0 ){ | |
| t = K > t ? K : t; | |
| }else{ | |
| return -1; | |
| } | |
| } | |
| return t; | |
| } | |
| let X = 5, A = []; | |
| A[0] = 1 | |
| A[1] = 3 | |
| A[2] = 1 | |
| A[3] = 4 | |
| A[4] = 2 | |
| A[5] = 3 | |
| A[6] = 5 | |
| A[7] = 4 | |
| console.log(solution(X, A)) ; // 6 |
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