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# rodcisal/getPairs.ts

Created May 2, 2021 09:29
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Get Pairs that makes target number
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 function getPairs (array: number[], target: number): number { const needed = {}; let result = 0; for (let i = 0; i < array.length; i++) { // store pair that makes target with element from original array and index needed[array[i] - target] = i; } for (let i = 0; i < array.length; i++) { // check that pair that makes target exists and it's not same element in original array if (needed[array[i]] !== undefined && needed[array[i]] !== i) { result++ // element used in pair so deleted from needed array needed[array[i]] = undefined; } } return result; } // Time complexity is O(N+N) which means O(N) // https://stackoverflow.com/questions/5872120/time-complexity-of-two-for-loops test('passing array [1, 2, 3, 4] with target 1', () => { const r = getPairs([1,2,3,4], 1); expect(r).toBe(3); }); test('passing array [-2, 2, 0, 4] with target 2', () => { const r = getPairs([-2, 2, 0, 4], 2); expect(r).toBe(3); }); test('passing array [-2, 2, 2, 0, 4] with target 2', () => { const r = getPairs([-2, 2, 2, 0, 4], 2); expect(r).toBe(3); }); test('passing array [2, 4, 6, 8] with target 2', () => { const r = getPairs([2, 4, 6, 8], 2); expect(r).toBe(3); }); test('passing array [8, 5, 3, 0] with target 3', () => { const r = getPairs([8, 5, 3, 0], 3); expect(r).toBe(2); }); test('with empty array', () => { const r = getPairs([], 10); expect(r).toBe(0); }); test('passing array [-2,2] with target 0', () => { const r = getPairs([-2,2],0); expect(r).toBe(0); });
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