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Updated to spec

Source Link

edited Apr 29, 2016 at 18:32

11.2k
1
25
50

JavaScript (Firefox), 80 bytes

solution=

a=>b=>[for(_ of(i=0,a))for(_ of(j=!++i,a))if(b.searchincludes(s=a.substr(j++,i))<0)s][0]

document.write("<pre>"+
[ [ "test", "best" ], [ "wes", "west" ], [ "red", "dress" ] ]
.map(c=>c+": "+solution(c[0])(c[1])).join`\n`)

Test works only in Firefox. Returns undefined if there is no substring.

JavaScript (Firefox), 80 bytes

solution=

a=>b=>[for(_ of(i=0,a))for(_ of(j=!++i,a))if(b.search(s=a.substr(j++,i))<0)s][0]

document.write("<pre>"+
[ [ "test", "best" ], [ "wes", "west" ], [ "red", "dress" ] ]
.map(c=>c+": "+solution(c[0])(c[1])).join`\n`)

Test works only in Firefox. Returns undefined if there is no substring.

JavaScript (Firefox), 80 bytes

solution=

a=>b=>[for(_ of(i=0,a))for(_ of(j=!++i,a))if(b.includes(s=a.substr(j++,i)))s][0]

document.write("<pre>"+
[ [ "test", "best" ], [ "wes", "west" ], [ "red", "dress" ] ]
.map(c=>c+": "+solution(c[0])(c[1])).join`\n`)

Test works only in Firefox. Returns undefined if there is no substring.

Source Link

created Apr 29, 2016 at 15:56

user81655

11.2k
1
25
50

JavaScript (Firefox), 80 bytes

solution=

a=>b=>[for(_ of(i=0,a))for(_ of(j=!++i,a))if(b.search(s=a.substr(j++,i))<0)s][0]

document.write("<pre>"+
[ [ "test", "best" ], [ "wes", "west" ], [ "red", "dress" ] ]
.map(c=>c+": "+solution(c[0])(c[1])).join`\n`)

Test works only in Firefox. Returns undefined if there is no substring.