Minimum Image Processing Cost

You must process n images. Image i requires a filter on every day from startDay[i] through endDay[i], inclusive, and filtering that image costs filterCost[i] per day.
An exclusive offer allows you to apply a filter to all n images for discountPrice on a day. Determine the minimum total cost of processing all images, and return that cost modulo 10^9 + 7.

Function
getMinProcessingCost(filterCost: int[], startDay: int[], endDay: int[], discountPrice: int) → int

Examples
Example 1
filterCost = [2,3,4]
startDay = [1,1,2]
endDay = [2,3,4]
discountPrice = 6
return = 21
For filterCost = [2, 3, 4], startDay = [1, 1, 2], endDay = [2, 3, 4], and discountPrice = 6, the individual filtering costs by day are:
Daily individual filtering costsDayImagesTotal cost
1[1, 2]2 + 3 = 5
2[1, 2, 3]2 + 3 + 4 = 9
3[2, 3]3 + 4 = 7
4[3]4
Using the all-images offer for 6 on days 2 and 3 is optimal. The final cost is 5 + 6 + 6 + 4 = 21, so its value modulo 10^9 + 7 is 21.

The trick: for each day, compute the sum of filterCost for images active that day, then add min(dailySum, discountPrice) to the total. Build the daily sums with a difference array: add filterCost[i] at startDay[i], subtract it at endDay[i]+1, then take a running prefix sum. The pitfall is the modulo. Take the min on the true values first, then reduce mod 10^9+7. Comparing already-reduced numbers gives wrong answers. Use 64-bit integers for sums, since costs times image counts can overflow 32 bits. Another trap is looping over every day for every image, which is O(n * range) and times out when day ranges are huge. If the ranges are very large, sort the event points instead of allocating an array. If your head goes blank mid-assessment, StealthCoder can supply the sweep as a hedge.

If this hits your live OA and you blank, StealthCoder solves it in seconds, invisible to the proctor.