Untitled
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javascript
a year ago
873 B
5
Indexable
function solution(numbers, k) { let count = 0; for (let start = 0; start < numbers.length; start++) { const freq = new Map(); for (let end = start; end < numbers.length; end++) { const num = numbers[end]; const prevCount = freq.get(num) || 0; freq.set(num, prevCount + 1); // Calculate the number of pairs for the current number let numPairs = Math.floor(prevCount / 2); // If we formed a new pair with the current number, check if we have k pairs in total if (prevCount % 2 === 1 && (prevCount + 1) % 2 === 0) { let totalPairs = Array.from(freq.values()).reduce((acc, val) => acc + Math.floor(val / 2), 0); if (totalPairs >= k) { count++; } } } } return count; }
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