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//In an array 1-100 exactly one number is duplicate how do you find it? | |
const findOneDuplicateNum = arr => { | |
if (!Array.isArray(arr) || !arr.length) { | |
return null; | |
} | |
let count = {}; | |
for (let i = 0; i < arr.length; i++) { | |
if (count[arr[i]]) { | |
return arr[i]; | |
} else { | |
count[arr[i]] = true; | |
} | |
} | |
return null; | |
}; | |
//Time-complexity: O(n) | |
//test cases | |
const testCase = [ | |
{ input: [], output: null }, | |
{ input: ["a", "b", "c"], output: null }, | |
{ input: "engimatic", output: null }, | |
{ input: [1, 2, 3, 4, 5, 2, 6], output: 2 }, | |
{ input: [1, 2, 3, 4, 5, 3, 6], output: 3 }, | |
{ input: [1, 2, 3, 4, 5, 6, 6], output: 6 }, | |
{ input: [1, 1, 3, 4, 5, 2, 6], output: 1 } | |
]; | |
testCase.forEach(({ input, output }, index) => { | |
console.log( | |
`TEST CASE:${index} (${input}):(${ | |
findOneDuplicateNum(input) === output ? "success" : "failure" | |
})` | |
); | |
}); |
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Perfect!