Minimum Equal Sum of Two Arrays After Replacing Zeros solution and explanation

def min_equal_sum(nums1, nums2):
    sum1, sum2 = 0, 0
    zeros1, zeros2 = 0, 0
    
    for num in nums1:
        if num == 0:
            zeros1 += 1
        else:
            sum1 += num
    
    for num in nums2:
        if num == 0:
            zeros2 += 1
        else:
            sum2 += num
    
    diff = abs(sum1 - sum2)
    if zeros1 + zeros2 < diff:
        return -1
    
    return sum1 + sum2 + max(zeros1, zeros2)

print(min_equal_sum([3, 2, 0, 1, 0], [6, 5, 0]))

class Solution {
    public int minEqualSum(int[] nums1, int[] nums2) {
        int sum1 = 0, sum2 = 0, zeros1 = 0, zeros2 = 0;
        for (int n : nums1) {
            if (n == 0) zeros1++;
            else sum1 += n;
        }
        for (int n : nums2) {
            if (n == 0) zeros2++;
            else sum2 += n;
        }
        int diff = Math.abs(sum1 - sum2);
        if (zeros1 + zeros2 < diff) return -1;
        return (sum1 + sum2 + ((sum1 > sum2) ? zeros1 : zeros2));
    }
    public static void main(String[] args) {
        Solution s = new Solution();
        int[] nums1 = {3, 2, 0, 1, 0};
        int[] nums2 = {6, 5, 0};
        System.out.println(s.minEqualSum(nums1, nums2));
    }
}

#include <iostream>
#include <vector>
#include <cmath>
using namespace std;

int minEqualSum(vector<int>& nums1, vector<int>& nums2) {
    int sum1 = 0, sum2 = 0, zeros1 = 0, zeros2 = 0;
    for (int n : nums1) {
        if (n == 0) zeros1++;
        else sum1 += n;
    }
    for (int n : nums2) {
        if (n == 0) zeros2++;
        else sum2 += n;
    }
    int diff = abs(sum1 - sum2);
    if (zeros1 + zeros2 < diff) return -1;
    return (sum1 + sum2 + ((sum1 > sum2) ? zeros1 : zeros2));
}

int main() {
    vector<int> nums1 = {3, 2, 0, 1, 0};
    vector<int> nums2 = {6, 5, 0};
    cout << minEqualSum(nums1, nums2) << endl;
    return 0;
}

function minEqualSum(nums1, nums2) {
    let sum1 = 0,
        sum2 = 0,
        zeros1 = 0,
        zeros2 = 0;
    for (let num of nums1) {
        if (num == 0) zeros1++;
        else sum1 += num;
    }
    for (let num of nums2) {
        if (num == 0) zeros2++;
        else sum2 += num;
    }
    let diff = Math.abs(sum1 - sum2);
    if (zeros1 + zeros2 < diff) return -1;
    return sum1 + sum2 + Math.max(zeros1, zeros2);
}

console.log(minEqualSum([3, 2, 0, 1, 0], [6, 5, 0]));

#include <stdio.h> 
#include <stdlib.h>

int min_equal_sum(int* nums1, int size1, int* nums2, int size2) {
    int sum1 = 0, sum2 = 0, zeros1 = 0, zeros2 = 0;
    
    for (int i = 0; i < size1; i++) {
        if (nums1[i] == 0) zeros1++;
        else sum1 += nums1[i];
    }
    for (int i = 0; i < size2; i++) {
        if (nums2[i] == 0) zeros2++;
        else sum2 += nums2[i];
    }

    int diff = abs(sum1 - sum2);
    if (zeros1 + zeros2 < diff) return -1;
    return (sum1 + sum2 + ((sum1 > sum2) ? zeros1 : zeros2));
}

int main() {
    int nums1[] = {3, 2, 0, 1, 0};
    int nums2[] = {6, 5, 0};
    int result = min_equal_sum(nums1, 5, nums2, 3);
    printf("%d\n", result);
    return 0;
}

using System;

class Solution {
    public int minEqualSum(int[] nums1, int[] nums2) {
        int sum1 = 0, sum2 = 0, zeros1 = 0, zeros2 = 0;
        foreach (int n in nums1) {
            if (n == 0) zeros1++;
            else sum1 += n;
        }
        foreach (int n in nums2) {
            if (n == 0) zeros2++;
            else sum2 += n;
        }
        int diff = Math.Abs(sum1 - sum2);
        if (zeros1 + zeros2 < diff) return -1;
        return (sum1 + sum2 + ((sum1 > sum2) ? zeros1 : zeros2));
    }
    
    static void Main() {
        Solution s = new Solution();
        int[] nums1 = {3, 2, 0, 1, 0};
        int[] nums2 = {6, 5, 0};
        Console.WriteLine(s.minEqualSum(nums1, nums2));
    }
}

# An illustrative Python concept would involve significant accounting for DP conditions.
# Commented here to reflect strategic limitations.

# Potential implementations outstrip the feasible given the exigencies of constraints and impact on time/space limits.

// Java also faces a similar challenge: efficient tracking and dynamic state management is demanding and complex.
/* The constraints suggest pursuit in practicality lies in other approaches due to scalability issues in DP efforts for such scale limits. */

// This comment placeholder suits for C++ as a similar evaluation of complexities with possibilities beyond practical efforts.

// JavaScript approaches comparative challenges given expansive DP states.
/* Using JavaScript flexibility extends potential time concerns beyond practical solution implementation amid input-scale considerations. */

// A Dynamic programming approach might involve tracking potential sums in a DP-like fashion.
/* Implementing it in C efficiently requires a careful strategy to adjust minimal sum changes and balance different possibilities, which involves significant complexity and is impractical for maximum constraints. */

// In C# such state management lands this approach cumbersome.
// Implementing as a DP strategy involves intensive memory and time management.

// Similar practical unfeasibility under the constraints provided.

function minSum(nums1: number[], nums2: number[]): number {
    let [s1, s2] = [0, 0];
    let hasZero = false;
    for (const x of nums1) {
        if (x === 0) {
            hasZero = true;
        }
        s1 += Math.max(x, 1);
    }
    for (const x of nums2) {
        s2 += Math.max(x, 1);
    }
    if (s1 > s2) {
        return minSum(nums2, nums1);
    }
    if (s1 === s2) {
        return s1;
    }
    return hasZero ? s2 : -1;
}

func minSum(nums1 []int, nums2 []int) int64 {
	s1, s2 := 0, 0
	hasZero := false
	for _, x := range nums1 {
		if x == 0 {
			hasZero = true
		}
		s1 += max(x, 1)
	}
	for _, x := range nums2 {
		s2 += max(x, 1)
	}
	if s1 > s2 {
		return minSum(nums2, nums1)
	}
	if s1 == s2 {
		return int64(s1)
	}
	if hasZero {
		return int64(s2)
	}
	return -1
}