Https://Www.Typescriptlang.Org/Play?#Code/MYewdgzgLgBAJgS2gJwQIwK5QKYBkmwC8MAFAFAyUwA2BAXDGBgLZrbIDaAugDQVVQAhsgDm2KAEkwcbAA8GTVuz5UYQ0eIBy2AO4A1QdQzYFLNsjIBKU0s5cYhAHwwA3v0oB6DzCRqAFtiMujAAbobGMCLI2II4yP6CYP6BINRwoeHYAPzuMKCQsEgA4tGx7A5qwmJQ2vqZMM600Bzq1VIyslxkuV6R4snwCABmQ+ZYYMCBbFA62OMDYMGJ6alwPd6aAPIAKgCiDFABgyMwzIIAnjBsQSKxCCHYPDAQIHngD8iFsFCvAA4gEAQUHu2Fy1H6iBOxAAsrE-AA6QRoCAkJpQFpVcTtOT2AC0lQ0NV0BiM2Es3VUvQg-VaNNeQIgQR0GVJYIIGMJ2M6FVpRLqrPWA0hQxg0V+0WpYCgEDB-QQ0jkFQADBSqDo-AhwaRhQ0YErLK5cpTvBAANYIX4DXkwbDg5jjQoK2RPF4+WA6cAAclg12YWDKa1UqmGpHlHQchGIvK5MEsbiDCZ8ToA1MmjYn8sCmKDEzAAL6qhO9CXqR3h-6A4HgHwi2YwDXfI5nWRJjrpyghkhhxWNAjw8FgESHA3x3NUbst4gqsdUTPy4zt-OFoPFUpxGBZGjsidcXH4hhojg71PL4MptO54V73IF3LRKAYZBJNFkW9DDATKtJHDQAAiBFQTAcHwaASBHXJ8mgNRsGgRliA4RdRxneVfiwBgOAAZieLCYAARl4RdVGjJ0GAAJhUGcCWqWoSWMBglQomcdESKB0IAFmwp4lS6XM8z4RDCPHMBUNYmBMKeDi9QIyiqKxEi8MYsdeRozIyMU3NmKldinnIvCeMTPjTyoJCxxQtCxJwyzpMo4iOno9TE2U4lVIUwTKE00SOFwiSJP0hMC1ULpBWAAJgFNRkMJtO0HUZEhzhiPwnj8YQ4FxUAZDgclVHfT8EGrJBdgARwwQxUQIAANGxzG4J40QATSq9huGsK4QFSGIkhM4MRTK6ByvhAArEB5RIAAiUaDUjYh6sG4awDGibwJk+9H2-ZAF14xcVqfGAhkMakb2XIYQHiEhIO+GCoAAYUEakYBAEUf2lJaM3AKCRBAIhBhQdAsDwAgSCem7qXhMyoCeIHbuweFbLkCHLuB6GnP5YxyUXTsAEICuK0qPvB6DoER+EPMsF6Z0g9rEQgalPhIPbqGpJ5RoAIUEOAJsXPMorurrXsgSnqBAEQxs2U0Oc21QC1vJ7-X+OD-TAywgA

Basically round-robin increment or decrement until it's done; it's a valid strategy depending on desired outcome (my approach is weighted, so if one candidate is 0 in original array, it'll never increase (unless all candidates were 0), there's value in that) And with your strategy, you could reduce iterations by doing Math.floor(available / (length - 1)) and increment/decrement as much as possible at a time. Also you can remove isgreater check by burning sign into the correction value

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